MCQ 1012 Marks
If three coterminous edges of a parallelopiped are represented by $a - b , b - c$ and $c - a$, then its volume is
- A
$[ a b c ]$
- B
$2[ a b c ]$
- C
$[ a b c ]^2$
- ✓
$0$
Answer(D) Volume of parallelopiped $=[a-b \ b-c \ c-a]$
$\begin{array}{l}=[\overline{ a } \overline{ b } \overline{ c }]-[\overline{ b } \overline{ c } \overline{ a }] \\ =[\overline{ a } \overline{ b } \overline{ c }]-[\overline{ a } \overline{ b } \overline{ c }] \\ =0\end{array}$
View full question & answer→MCQ 1022 Marks
If the volume of the parallelopiped with coterminus edges $-p \hat{i}+5 k, \hat{i}-\hat{j}+q \hat{k}$ and $3 \hat{i}-5 \hat{j}$ is 8 , then .
- ✓
$5 p q+18=0$
- B
$3 p q-18=0$
- C
$p q+18=0$
- D
$p q-18=0$
AnswerCorrect option: A. $5 p q+18=0$
(A) Volume of parallelopiped $=\left|\begin{array}{rrr}- p & 0 & 5 \\ 1 & -1 & q \\ 3 & -5 & 0\end{array}\right|=8$
$\begin{array}{l}\Rightarrow-p(0+5 q)+5(-5+3)=8 \\ \Rightarrow-5 p q-18=0 \\ \Rightarrow 5 p q+18=0\end{array}$
View full question & answer→MCQ 1032 Marks
The volume of parallelopiped with vector $\bar{a}+2 \bar{b}-\bar{c}, \bar{a}-\bar{b}$ and $\bar{a}-\bar{b}-\bar{c}$ is equal to $k[\bar{a} \bar{b} \bar{c}]$, then $k=$
Answer(B) Volume of parallelopiped
$=\left|\begin{array}{ccc}1 & 2 & -1 \\ 1 & -1 & 0 \\ 1 & -1 & -1\end{array}\right|[\overline{ a } \overline{ b } \overline{ c }]= k [\overline{ a } \overline{ b } \overline{ c }]$
$\begin{array}{l}\Rightarrow 1(1-0)-2(-1-0)-1(-1+1)= k \\ \Rightarrow 1+2-0= k \Rightarrow k =3\end{array}$
View full question & answer→MCQ 1042 Marks
If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors and $\lambda$ is a real number then $\left[\lambda(\bar{a}+\bar{b}) \ \lambda^2 \bar{b} \ \lambda \bar{c}\right]=[\bar{a} \bar{b}+\bar{c} \bar{b}]$ for
- A
exactly three values of $\lambda$
- B
exactly two values of $\lambda$
- C
exactly one value of $\lambda$
- ✓
no value of $\lambda$
AnswerCorrect option: D. no value of $\lambda$
(D) $\left[\lambda(\overline{ a }+\overline{ b }) \lambda^2 \overline{b} \quad \lambda \overline{ c }\right]=\left[\begin{array}{lll}\overline{ a } & \overline{ b }+\overline{ c } & \overline{ b }\end{array}\right]$
$\Rightarrow \lambda^4[\overline{ a }+\overline{ b } \overline{ b } \overline{ c }]=[\overline{ a } \overline{ b }+\overline{ c } \overline{ b }]$
$\begin{array}{l}\Rightarrow \lambda^4\{[\overline{ abc }]+[\overline{ b } \overline{ bc }]\}=\{[\overline{ a } \overline{ bc }]+[\overline{ ac } \overline{ b }]\} \\ \Rightarrow \lambda^4[\overline{ a } \overline{ bc }]=-[\overline{ a } \overline{ b } \overline{ c }]\end{array}$
$\Rightarrow\left(\lambda^4+1\right)[\overline{ a } \overline{ bc }]=0$
But, $[\overline{ a } \overline{ b } \overline{ c }] \neq 0$.
$\therefore \quad \lambda^4+1=0$
This is not true for any real value of $\lambda$.
View full question & answer→MCQ 1052 Marks
If $\bar{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k}), \bar{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$, then the value of $(2 \bar{a}-\bar{b}) \cdot\{(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b})\}$ is
Answer(A) Since $\bar{a} \cdot \bar{b}=0$
$\therefore \quad \overline{ a }$ and $\overline{ b }$ are perpendicular unit vectors.
Now, $(2 \overline{ a }-\overline{ b }) \cdot\{(\overline{ a } \times \overline{ b }) \times(\overline{ a }+2 \overline{b})\}$
$\begin{array}{l}=[2 \overline{ a }-\overline{ b } \overline{ a } \times \overline{ b } \overline{ a }+2 \overline{b}] \\ =-[\overline{ a } \times \overline{ b } 2 \overline{ a }-\overline{ b } \overline{ a }+2 \overline{b}] \\ =-(\overline{ a } \times \overline{ b }) \cdot\{(2 \overline{ a }-\overline{ b }) \times(\overline{ a }+2 \overline{b})\}\end{array}$
$=-(\overline{ a } \times \overline{ b }) \cdot 5(\overline{ a } \times \overline{ b })$
$=-5|\overline{ a } \times \overline{ b }|=-5|\overline{ a }|^2|\overline{b}|^2$ $\ldots .[\because \bar{a} \perp \bar{b}]$
$=-5$ $\ldots \ldots[\because|\overline{ a }|=|\overline{ b }|=1]$
View full question & answer→MCQ 1062 Marks
$[\bar{a}+2 \bar{b}-\bar{c} \quad \bar{a}-\bar{b} \quad \bar{a}-\bar{b}-\bar{c}]=$
- A
$\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$
- ✓
$3\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$
- C
$0$
- D
$2\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$
AnswerCorrect option: B. $3\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$
(B) $[\overline{ a }+2 \overline{b}-\overline{ c } \quad \overline{ a }-\overline{ b } \quad \overline{ a }-\overline{ b }-\overline{ c }]$
$=(\overline{ a }+2 \overline{b}-\overline{ c }) \cdot\{(\overline{ a }-\overline{ b }) \times(\overline{ a }-\overline{ b }-\overline{ c })\}$
$=(\overline{ a }+2 \overline{b}-\overline{ c })$ $\cdot\{\overline{ a } \times \overline{ a }-\overline{ a } \times \overline{ b }-\overline{ a } \times \overline{ c }-\overline{ b } \times \overline{ a }+\overline{ b } \times \overline{ b }+\overline{ b } \times \overline{ c }\}$
$=(\overline{ a }+2 \overline{b}-\overline{ c })\{\overline{ b } \times \overline{ a }-\overline{ a } \times \overline{ c }-\overline{ b } \times \overline{ a }+\overline{ b } \times \overline{ c }\}$
$=(\overline{ a }+2 \overline{b}-\overline{ c })\{-\overline{ a } \times \overline{ c }+\overline{ b } \times \overline{ c }\}$
$\begin{array}{l}=\left[\begin{array}{lll}\overline{ a } & \overline{ b } & \overline{ c }\end{array}\right]-2\left[\begin{array}{lll}\overline{ b } & \overline{ a } & \overline{ c }\end{array}\right] \\ =\left[\begin{array}{lll}\overline{ a } & \overline{ b } & \overline{ c }\end{array}\right]+2\left[\begin{array}{lll}\overline{ a } & \overline{ b } & \overline{ c }\end{array}\right] \\ =3\left[\begin{array}{lll}\overline{ a } & \overline{ b } & \overline{ c }\end{array}\right]\end{array}$
View full question & answer→MCQ 1072 Marks
If $\overline{ a }, \overline{ b }, \overline{ c }$ be any three non-coplanar vectors, then $[\bar{a}+\bar{b} \ \bar{b}+\bar{c} \ \bar{c}+\bar{a}]=$
- A
$|\bar{a} \bar{b} \bar{c}|$
- ✓
$2[\bar{a} \bar{b} \bar{c}]$
- C
$[\overline{ a } \overline{ b } \overline{ c }]^2$
- D
$2[\bar{a} \bar{b} \bar{c}]^2$
AnswerCorrect option: B. $2[\bar{a} \bar{b} \bar{c}]$
(B) $[\overline{ a }+\overline{ b } \ \overline{ b }+\overline{ c } \ \overline{ c }+\overline{ a }]$
$=(\overline{ a }+\overline{ b }) \cdot\{(\overline{ b }+\overline{ c }) \times(\overline{ c }+\overline{ a })\}$
$=(\overline{ a }+\overline{ b }) \cdot(\overline{ b } \times \overline{ c }+\overline{ b } \times \overline{ a }+\overline{ c } \times \overline{ c }+\overline{ c } \times \overline{ a })$
$=(\overline{ a }+\overline{ b }) \cdot(\overline{ b } \times \overline{ c }+\overline{ b } \times \overline{ a }+\overline{ c } \times \overline{ a })$ $\ldots[\because \bar{c} \times \bar{c}=0]$
$=\overline{ a } \cdot(\overline{ b } \times \overline{ c })+\overline{ a } \cdot(\overline{ b } \times \overline{ a })+\overline{ a } \cdot(\overline{ c } \times \overline{ a })$ $+\overline{ b } \cdot(\overline{ b } \times \overline{ c })+\overline{ b } \cdot(\overline{ b } \times \overline{ a })+\overline{ b } \cdot(\overline{ c } \times \overline{ a })$
$=[\overline{ a } \overline{ b } \overline{ c }]+[\overline{ b } \overline{ c } \overline{ a }]=2[\overline{ a } \overline{ b } \overline{ c }]$
View full question & answer→MCQ 1082 Marks
The value of $[\bar{a}-\bar{b} \ \bar{b}-\bar{c} \ \bar{c}-\bar{a}]$, where $|\bar{a|}=1,|\bar{b}|=5$ and $| \bar{c} \mid=3$ is
Answer(A) $\left[\begin{array}{llll}\overline{ a }-\overline{ b } & \overline{ b }-\overline{ c } & \overline{ c }-\overline{ a }\end{array}\right]$
$\begin{array}{l}=\{(\overline{ a }-\overline{ b }) \times(\overline{ b }-\overline{ c })\} \cdot(\overline{ c }-\overline{ a }) \\ =(\overline{ a } \times \overline{ b }-\overline{ a } \times \overline{ c }-\overline{ b } \times \overline{ b }+\overline{ b } \times \overline{ c }) \cdot(\overline{ c }-\overline{ a }) \\ =(\overline{ a } \times \overline{ b }+\overline{ c } \times \overline{ a }+\overline{ b } \times \overline{ c }) \cdot(\overline{ c }-\overline{ a })\end{array}$
$=(\overline{ a } \times \overline{ b }) \cdot \overline{ c }-(\overline{ a } \times \overline{ b }) \cdot \overline{ a }+(\overline{ c } \times \overline{ a }) \cdot \overline{ c }$ $-(\overline{ c } \times \overline{ a }) \cdot \overline{ a }+(\overline{ b } \times \overline{ c }) \cdot \overline{ c }-(\overline{ b } \times \overline{ c }) \cdot \overline{ a }$
$=\left[\begin{array}{lll}\overline{ a } & \overline{ b } & \overline{ c }\end{array}\right]-\left[\begin{array}{lll}\overline{ a } & \overline{ b } & \overline{ a }\end{array}\right]+\left[\begin{array}{lll}\overline{ c } & \overline{ a } & \overline{ c }\end{array}\right]-\left[\begin{array}{lll}\overline{ c } & \overline{ a } & \overline{ a }\end{array}\right]$ $+\left[\begin{array}{lll}\overline{ b } & \overline{ c } & \overline{ c }\end{array}\right]-\left[\begin{array}{lll}\overline{ b } & \overline{ c } & \overline{ a }\end{array}\right]=0$
View full question & answer→MCQ 1092 Marks
If $\bar{a}, \bar{b}$ and $\bar{c}$ are three non-coplanar vectors then $(\bar{a}+\bar{b}+\bar{c}) \cdot[(\bar{a}+\bar{b}) \times(\bar{a}+\bar{c})]=$
- A
$0$
- B
$[\overline{ a } \overline{ b } \overline{ c }]$
- ✓
$-[\overline{ a } \overline{ b } \overline{ c }]$
- D
$2[\overline{ a }, \overline{ b }, \overline{ c }]$
AnswerCorrect option: C. $-[\overline{ a } \overline{ b } \overline{ c }]$
(C) $[\overline{ a }+\overline{ b }+\overline{ c }] \cdot[(\overline{ a }+\overline{ b }) \times(\overline{ a }+\overline{ c })]$
$=(\bar{a}+\bar{b}) \cdot[(\bar{a}+\bar{b}) \times(\bar{a}+\bar{c})]$ $+\bar{c} \cdot[(\bar{a}+\bar{b}) \times(\bar{a}+\bar{c})]$
$=0+[\overline{ c } \overline{ a }+\overline{ b } \overline{ a }+\overline{ c }]$
$=[\overline{ c } \overline{ a } \overline{ a }+\overline{ c }]+[\overline{ c } \overline{ b } \overline{ a }+\overline{ c }]$
$=\left[\begin{array}{lll}\bar{c} & \bar{a} & \bar{a}\end{array}\right]+\left[\begin{array}{lll}\bar{c} & \bar{a} & \bar{c}\end{array}\right]+\left[\begin{array}{lll}\bar{c} & \bar{b} & \bar{a}\end{array}\right]+\left[\begin{array}{lll}\bar{c} & \bar{b} & \bar{c}\end{array}\right]$
$\begin{array}{l}=0+0+[\overline{ c } \overline{ b } \overline{ a }]+0 \\ =-[\overline{ a } \overline{ b } \overline{ c }]\end{array}$
View full question & answer→MCQ 1102 Marks
If $\overline{ u }, \overline{ v }$ and $\overline{ w }$ are three non-coplanar vectors, then $(\bar{u}+\bar{v}-\bar{w}) \cdot[(\bar{u}-\bar{v}) \times(\bar{v}-\bar{w})]$ equals
- A
$0$
- ✓
$\bar{u} \cdot(\bar{v} \times \bar{w})$
- C
$\bar{u} \cdot(\bar{w} \times \bar{v})$
- D
$3 \overline{ u } \cdot(\overline{ v } \times \overline{ w })$
AnswerCorrect option: B. $\bar{u} \cdot(\bar{v} \times \bar{w})$
(B) $(\bar{u}+\bar{v}-\bar{w}) \cdot[(\bar{u}-\bar{v}) \times(\bar{v}-\bar{w})]$
$=\overline{ u } \cdot(\overline{ u } \times \overline{ v })-\overline{ u } \cdot(\overline{ u } \times \overline{ w })+\overline{ u } \cdot(\overline{ v } \times \overline{ w })+\overline{ v } \cdot(\overline{ u } \times \overline{ v }) -\overline{ v } \cdot(\overline{ u } \times \overline{ w })+\overline{ v } \cdot(\overline{ v } \times \overline{ w })-\overline{ w } \cdot(\overline{ u } \times \overline{ v })$ $+\bar{w} \cdot(\bar{u} \times \bar{w})-\bar{w} \cdot(\bar{v} \times \bar{w})$
$\begin{array}{l}=[\overline{ u } \overline{ v } \overline{ w }]-[\overline{ v } \overline{ u } \overline{ w }]-[\overline{ w } \overline{ u } \overline{ v }] \\ =[\overline{ u } \overline{ v } \overline{ w }]+[\overline{ u } \overline{ v } \overline{ w }]-[\overline{ u } \overline{ v } \overline{ w }]=\overline{ u } \cdot(\overline{ v } \times \overline{ w })\end{array}$
View full question & answer→MCQ 1112 Marks
$(\bar{a}+\bar{b}) \cdot(\bar{b}+\bar{c}) \times(\bar{a}+\bar{b}+\bar{c})=$
- A
-$[\bar{a} \bar{b} \bar{c}]$
- ✓
$[\bar{a} \bar{b} \bar{c}]$
- C
$0$
- D
$2[\overline{ a } \overline{ b } \overline{ c }]$
AnswerCorrect option: B. $[\bar{a} \bar{b} \bar{c}]$
(B) $(\overline{ a }+\overline{ b }) \cdot(\overline{ b }+\overline{ c }) \times(\overline{ a }+\overline{ b }+\overline{ c })$
$=(\overline{ a }+\overline{ b }) \cdot[\overline{ b } \times \overline{ a }+\overline{ b } \times \overline{ c }+\overline{ c } \times \overline{ a }+\overline{ c } \times \overline{ b }]$
$=[\overline{ a } \overline{ b } \overline{ a }]+[\overline{ a } \overline{ b } \overline{ c }]+[\overline{ a c a }]+[\overline{ a c} \overline{b}]$ $+[\overline{ b } \overline{ b } \overline{ a }]+[\overline{ b } \overline{ b } \overline{ c }]+[\overline{ b } \overline{ c } \overline{ a }]+[\overline{ b } \overline{ c } \overline{ b }]$
$=0+[\overline{ a } \overline{ b } \overline{ c }]+0+[\overline{ a } \overline{ c } \overline{ b }]+0+0+[\overline{ b } \overline{ c } \overline{ a }]+0$
$=[\overline{ a } \overline{ b } \overline{ c }]-[\overline{ a } \overline{ b } \overline{ c }]+[\overline{ a } \overline{ b } \overline{ c }]=[\overline{ a } \overline{ b } \overline{ c }]$
View full question & answer→MCQ 1122 Marks
For any three vectors $\bar{a}, \bar{b}$ and $\bar{c}$, $(\bar{a}-\bar{b}) \cdot[(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})]$ is equal to :
- A
$2 \bar{a} \cdot(\bar{b} \times \bar{c})$
- B
$[\bar{a} \bar{b} \bar{c}]$
- C
$[\overline{ a } \overline{ b } \overline{ c }]^2$
- ✓
$0$
Answer(D) $(\bar{a}-\bar{b}) \cdot[(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})]$
$=(\bar{a}-\bar{b})$ $\cdot[\overline{ b } \times \overline{ c }+\overline{ b } \times \overline{ a }+\overline{ c } \times \overline{ c }+\overline{ c } \times \overline{ a }]$
$=\bar{a} \cdot(\bar{b} \times \bar{c})+\bar{a} \cdot(\bar{b} \times \bar{a})+\bar{a} \cdot(\bar{c} \times \bar{a})$ $-\overline{ b } \cdot(\overline{ b } \times \overline{ c })-\overline{ b } \cdot(\overline{ b } \times \overline{ a })-\overline{ b } \cdot(\overline{ c } \times \overline{ a })$
$=[\overline{ a } \overline{ b } \overline{ c }]-[\overline{ a } \overline{ b } \overline{ c }]=0$
View full question & answer→MCQ 1132 Marks
$\overline{ a } \cdot[(\overline{ b }+\overline{ c }) \times(\overline{ a }+\overline{ b }+\overline{ c })]$ is equal to
- A
$\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$
- B
$2[\bar{a} \bar{b} \bar{c}]$
- C
$3[\bar{a} \bar{b} \bar{c}]$
- ✓
$0$
Answer(D) $\overline{ a } \cdot[(\overline{ b }+\overline{ c }) \times(\overline{ a }+\overline{ b }+\overline{ c })]$
$=\overline{ a } \cdot[\overline{ b } \times \overline{ a }+\overline{ b } \times \overline{ c }+\overline{ c } \times \overline{ a }+\overline{ c } \times \overline{ b }]$ $[\because \overline{ b } \times \overline{ b }=0, \overline{ c } \times \overline{ c }=0]$
$\begin{array}{l}=[\overline{ a } \overline{ b } \overline{ a }]+[\overline{ a } \overline{ b } \overline{ c }]+[\overline{ a } \overline{ c } \overline{ a }]+[\overline{ a } \overline{ c } \overline{ b }] \\ =0+[\overline{ a } \overline{ b } \overline{ c }]+0-[\overline{ a } \overline{ b } \overline{ c }] \\ =0\end{array}$
View full question & answer→MCQ 1142 Marks
If $\bar{a}, \bar{b}, \bar{c}$ are the three non-coplanar vectors and $\bar{p}, \bar{q}, \bar{r}$ are defined by the relations $\bar{p}=\frac{\bar{b} \times \bar{c}}{[\bar{a} \bar{b} \bar{c}]}, \bar{q}=\frac{\bar{c} \times \bar{a}}{[\bar{a} \bar{b} \bar{c}]}, \bar{r}=\frac{\bar{a} \times \bar{b}}{[\bar{a} \bar{b} \bar{c}]}$ then $(\bar{a}+\bar{b}) \cdot \bar{p}+(\bar{b}+\bar{c}) \cdot \bar{q}+(\bar{c}+\bar{a}) \cdot \bar{r}=$
Answer(D) $\overline{ p } \cdot(\overline{ a }+\overline{ b })=\overline{ p } \cdot \overline{ a }+\overline{ p } \cdot \overline{ b }$
$=\frac{(\overline{ b } \times \overline{ c }) \cdot \overline{ a }}{[\overline{ a } \overline{ b } \overline{ c }]}+\frac{(\overline{ b } \times \overline{ c }) \cdot \overline{ b }}{[\overline{ a } \overline{ b } \overline{ c }]}$
$=\frac{[\overline{ b } \overline{ c } \overline{ a }]}{[\overline{ a } \overline{ b } \overline{ c }]}+\frac{[\overline{ b } \overline{ c } \overline{ b }]}{[\overline{ a } \overline{ b } \overline{ c }]}$
$\begin{array}{l}=1+0 \\ =1\end{array}$
Similarly, $\overline{ q } \cdot(\overline{ b }+\overline{ c })=1$ and $\overline{ r } \cdot(\overline{ a }+\overline{ c })=1$
$\begin{array}{l}(\bar{a}+\bar{b}) \cdot \bar{p}+(\bar{b}+\bar{c}) \cdot \bar{q}+(\bar{c}+\bar{a}) \cdot \bar{r} \\ =1+1+1=3\end{array}$
View full question & answer→MCQ 1152 Marks
If $\overline{ p }=\frac{\overline{ b } \times \overline{ c }}{[\overline{ a } \overline{ b }\overline {c} ]}, \overline{ q }=\frac{\overline{ c } \times \overline{ a }}{[ \overline{a} \overline{ b } \overline{c} ]}, \overline{ r }=\frac{\overline{ a } \times \overline{ b }}{[\overline {a} \overline{ b } \overline{c} ]}$. where $\bar{a}$, $\bar{b}, \bar{c}$ are three non-coplanar vectors, then the value of $(\bar{a}+\bar{b}+\bar{c}) \cdot(\bar{p}+\bar{q}+\bar{r})$ is given by
Answer(A) $\overline{ p }+\overline{ q }+\overline{ r }=\frac{\overline{ b } \times \overline{ c }+\overline{ c } \times \overline{ a }+\overline{ a } \times \overline{ b }}{[\overline{ a } \overline{ b } \overline{ c }]}$
$(\overline{ a }+\overline{ b }+\overline{ c }) \cdot(\overline{ p }+\overline{ q }+\overline{ r })$
$=\frac{[\overline{ a } \overline{ b } \overline{ c }]+[\overline{ b } \overline{ c } \overline{ a }]+[\overline{ c } \overline{ a } \overline{ b }]}{[\overline{ a } \overline{ b } \overline{ c }]}$
$=3$
View full question & answer→MCQ 1162 Marks
If $\bar{a}, \bar{b}$ and $\bar{c}$ are non-coplanar, then the value of $\bar{a} \cdot\left\{\frac{\bar{b} \times \bar{c}}{3 \bar{b} \cdot(\bar{c} \times \bar{a})}\right\}-\bar{b} \cdot\left\{\frac{\bar{c} \times \bar{a}}{2 \bar{c} \cdot(\bar{a} \times \bar{b})}\right\}$ is
- A
$\frac{-1}{2}$
- B
$\frac{-1}{3}$
- ✓
$\frac{-1}{6}$
- D
$\frac{1}{6}$
AnswerCorrect option: C. $\frac{-1}{6}$
(C) $\overline{ a }, \overline{ b }$ and $\overline{ c }$ are non-coplanar.
So, $[\overline{ a } \overline{ bc }] \neq 0$
$\overline{ a } \cdot\left\{\frac{\overline{ b } \times \overline{ c }}{3 \overline{b} \cdot(\overline{ c } \times \overline{ a })}\right\}-\overline{ b } \cdot\left\{\frac{\overline{ c } \times \overline{ a }}{2 \overline{ c } \cdot(\overline{ a } \times \overline{ b })}\right\}$
$=\frac{[\overline{ a }\overline{ b }\overline{ c }]}{3[\overline{ b }\overline{ c }\overline{ a }]}-\frac{[\overline{ b }\overline{ c }\overline{ a }]}{2[\overline{ c }\overline{ a }\overline{ b }]}=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}$
View full question & answer→MCQ 1172 Marks
$\overline{ r }=l(\overline{b} \times \overline{ c })+ m (\overline{ c } \times \overline{ a })+ n (\overline{ a } \times \overline{ b })$ and $[\overline{ a } \overline{ b } \overline{ c }]=2$ then the value of $l+ m + n$ is
- A
$(\bar{a}+\bar{b}+\bar{c}) \cdot \bar{r}$
- ✓
$\frac{1}{2}(\bar{a}+\bar{b}+\bar{c}) \cdot \bar{r}$
- C
$\frac{1}{3}(\bar{a}+\bar{b}+\bar{c}) \cdot(\bar{a}+\bar{b}+\bar{c})$
- D
$\frac{2}{3}(\overline{ a }+\overline{ b }+\overline{ c }) \cdot \overline{ r }$
AnswerCorrect option: B. $\frac{1}{2}(\bar{a}+\bar{b}+\bar{c}) \cdot \bar{r}$
(B) $\overline{ r }=l(\overline{b} \times \overline{ c })+ m (\overline{ c } \times \overline{ a })+ n (\overline{ a } \times \overline{ b })$
$\therefore \quad \overline{ a } \cdot \overline{ r }=l_{ a } \cdot(\overline{ b } \times \overline{ c })+ m \overline{ a } \cdot(\overline{ c } \times \overline{ a })+ n \overline{ a } \cdot(\overline{ a } \times \overline{ b })$
$=l[\overline{ a } \overline{ b } \overline{ c }]+0+0$
$\overline{ a } \cdot \overline{ r }=2 l$ $\ldots$.(i) $[\because[\overline{ a } \overline{ b } \overline{ c }]=2]$
Similarly,
$\overline{ b } \cdot \overline{ r }=2 m$, ...(ii)
$\overline{ c .} \overline{ r }=2 n$ ...(iii)
∴ On adding equations (i), (ii) and (iii) we get $(\overline{ a }+\overline{ b }+\overline{ c }) \cdot \overline{ r }=2(l+ m + n )$
$\therefore \quad l+ m + n =\frac{1}{2}(\overline{ a }+\overline{ b }+\overline{ c }) \cdot \overline{ r }$
View full question & answer→MCQ 1182 Marks
If $a , b , c$ are non-coplanar vectors and $d =\lambda a +\mu b +\nu c$ then $\lambda$ is equal to
- A
$\frac{[ dbc ]}{[ bac ]}$
- ✓
$\frac{[b c d]}{[b c a]}$
- C
$\frac{[ bdc ]}{[ abc ]}$
- D
$\frac{[ cbd ]}{[ abc ]}$
AnswerCorrect option: B. $\frac{[b c d]}{[b c a]}$
(B) Since $d=\lambda a+\mu b+v c$
$\therefore \quad d \cdot(b \times c)=\lambda a \cdot(b \times c)+\mu b \cdot(b \times c)+v c \cdot(b \times c)$
$\Rightarrow d \cdot( b \times c )=\lambda[ a b c ]$
$\Rightarrow \lambda=\frac{[ dbc ]}{[ abc ]}=\frac{[ bcd ]}{[ bca ]}$
View full question & answer→MCQ 1192 Marks
Given vectors $\bar{a}, \bar{b}, \bar{c}$ such that $\bar{a} \cdot(\bar{b} \times \bar{c})=\lambda \neq 0$ the value of $\frac{(\bar{b} \times \bar{c}) \cdot(\bar{a}+\bar{b}+\bar{c})}{\lambda}$ is
- A
- ✓
- C
$-3 \lambda$
- D
$\frac{3}{\lambda}$
Answer(B) $\frac{(\overline{ b } \times \overline{ c }) \cdot(\overline{ a }+\overline{ b }+\overline{ c })}{\lambda}$
$=\frac{(\overline{ b } \times \overline{ c }) \cdot \overline{ a }+(\overline{ b } \times \overline{ c }) \cdot \overline{ b }+(\overline{ b } \times \overline{ c }) \cdot \overline{ c }}{\lambda}$
$=\frac{(\overline{ b } \times \overline{ c }) \cdot \overline{ a }+\overline{0}+\overline{0}}{\lambda}$
$=\frac{\overline{ a } \cdot(\overline{ b } \times \overline{ c })}{\lambda}=\frac{\lambda}{\lambda}=1$
View full question & answer→MCQ 1202 Marks
The number of distinct real values of $\lambda$ for which the vectors $\bar{a}=\lambda^3 \hat{i}+\hat{k}, \bar{b}=\hat{i}-\lambda^3 \hat{j}$ and $\overline{ c }=\hat{ i }+(2 \lambda-\sin \lambda) \hat{ j }-\lambda \hat{ k }$ are coplanar is
Answer(B) The given vectors are coplanar
$\therefore\left|\begin{array}{ccc}\lambda^3 & 0 & 1 \\ 1 & -\lambda^3 & 0 \\ 1 & 2 \lambda-\sin \lambda & -\lambda\end{array}\right|=0$
$\Rightarrow \lambda^3\left(\lambda^4-0\right)+1\left(2 \lambda-\sin \lambda+\lambda^3\right)=0$
$\Rightarrow \lambda^7+\lambda^3+2 \lambda=\sin \lambda$ ...(i)
This is true for $\lambda=0$.
For non-zero values of $\lambda$, equation (i) is $\lambda^6+\lambda^2+2=\frac{\sin \lambda}{\lambda}$ ...(ii)
We know that $\frac{\sin x}{x}<1$ for all $x \neq 0$.
∴ L.H.S. of (ii) is greater than 2 and R.H.S. is less than 1 .
So, (ii) is not true for any non-zero $\lambda$.
Hence, there is only one value of $\lambda$.
View full question & answer→MCQ 1212 Marks
If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors and $\lambda$ is a real number, then the vectors $\bar{a}+2 \bar{b}+3 \bar{c}, \lambda \bar{b}+4 \bar{c}$ and $(2 \lambda-1) \bar{c}$ are non-coplanar for
AnswerCorrect option: C. all except two values of $\lambda$
(C) Let $\bar{\alpha}, \bar{\beta}$ and $\bar{\gamma}$ be the given vectors $\bar{\alpha}, \bar{\beta}$ and $\bar{\gamma}$ are coplanar
$\therefore\left|\begin{array}{lcc}1 & 2 & 3 \\ 0 & \lambda & 4 \\ 0 & 0 & (2 \lambda-1)\end{array}\right|=0$
$\Rightarrow \lambda(2 \lambda-1)=0 \Rightarrow \lambda=0, \frac{1}{2}$
Hence, $\bar{\alpha}, \bar{\beta}, \bar{\gamma}$ are non-coplanar for all values of $\lambda$ except 0 and $\frac{1}{2}$.
View full question & answer→MCQ 1222 Marks
The number of distinct real values of $\lambda$, for which the vectors $-\lambda^2 \hat{i}+\hat{j}+\hat{k}, \hat{i}-\lambda^2 \hat{j}+\hat{k}$ and $\hat{ i }+\hat{ j }-\lambda^2 \hat{ k }$ are coplanar, is
Answer(C) Let $\overline{ a }, \overline{ b }$ and $\overline{ c }$ be the given vectors. The vectors are coplanar.
$\therefore\left|\begin{array}{ccc}-\lambda^2 & 1 & 1 \\ 1 & -\lambda^2 & 1 \\ 1 & 1 & -\lambda^2\end{array}\right|=0$
$\Rightarrow-\lambda^2\left(\lambda^4-1\right)-1\left(-\lambda^2-1\right)+1\left(1+\lambda^2\right)=0 \\ \Rightarrow \lambda^6-3 \lambda^2-2=0 \\ \Rightarrow\left(1+\lambda^2\right)^2\left(\lambda^2-2\right)=0 \\ \Rightarrow \lambda= \pm \sqrt{2}$
View full question & answer→MCQ 1232 Marks
If $\bar{a}, \bar{b}, \bar{c}$ are non-zero, non-collinear vectors, then the vectors $\bar{a}$ - $\bar{b}$ + $\bar{c}$, 4$\bar{a}$ - 7$\bar{b}$ - $\bar{c}$ and 3$\bar{a}$ + 6$\bar{b}$ + 6$\bar{c}$ are
- A
- B
- C
both collinear and co-planar.
- ✓
neither collinear nor coplanar
AnswerCorrect option: D. neither collinear nor coplanar
(D) Let $P (\overline{ p }), Q (\overline{ q }), R (\overline{ r })$ be the three points.
$\begin{aligned} \therefore & \overline{ p }=\overline{ a }-\overline{ b }+\overline{ c }, \overline{ q }=4 \overline{ a }-7 \overline{b}-\overline{ c } \text { and } \\ & \overline{ r }=3 \overline{ a }+6 \overline{b}+6 \overline{ c }\end{aligned}$
$\overline{ PQ }$ is not scalar multiple of $\overline{ PR }$
∴ they are not collinear.
$\left[\begin{array}{ccc}\bar{p} & \bar{q} & \bar{r}\end{array}\right]=\left|\begin{array}{ccc}1 & -1 & 1 \\ 4 & -7 & -1 \\ 3 & 6 & 6\end{array}\right|=36 \neq 0$
∴ they are not coplanar.
View full question & answer→MCQ 1242 Marks
Let $\bar{A}=\hat{i}+\hat{j}+\hat{k}, \bar{B}=\hat{i}, \bar{C}=C_1 \hat{i}+C_2 \hat{j}+C_3 \hat{k}$.
If $C _2=-1$, and $C _3=1$, then to make three vectors coplanar
AnswerCorrect option: D. No value of $C _1$ can be found
(D) Here $\overline{ C }= C _1 \hat{ i }-\hat{ j }+\hat{ k }$
To make three vectors coplanar, $[\overline{ A } \overline{ B } \overline{ C }]=0$
$\Rightarrow\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 0 & 0 \\ C_1 & -1 & 1\end{array}\right|=0$
$\Rightarrow 1(0-0)-1(1-0)+1(-1-0)=0$
∴ The value of $[\overline{ A } \overline{ B } \overline{ C }]$ is independent of $C _1$. Hence, no value of $C _1$ can be found.
View full question & answer→MCQ 1252 Marks
A vector perpendicular to $2 \hat{i}+\hat{j}+\hat{k}$ and coplanar with $\hat{i}+2 \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+2 \hat{k}$ is
AnswerCorrect option: A. $5(\hat{ j }-\hat{ k })$
(A) Let the vector be $a \hat{i}+b \hat{j}+c \hat{k}$.
It is perpendicular to $2 \hat{i}+\hat{j}+\hat{k}$.
$\therefore \quad 2 a+b+c=0$ ...(i)
The vector is coplanar with $\hat{i}+2 \hat{j}+\hat{k}$ and $\hat{ i }+\hat{ j }+2 \hat{ k }$
$\therefore\left|\begin{array}{ccc}a & b & c \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{array}\right|=0$
$\therefore \quad 3 a-b-c=0$ ...(ii)
On solving (i) and (ii), we get
$a=0, b=5, c=-5$
∴ The required vector is $5(\hat{ j }-\hat{ k })$
View full question & answer→MCQ 1262 Marks
If the given vectors $\left(-b c, b^2+b c, c^2+b c\right) \left( a ^2+ ac ,- ac , c ^2+ ac \right)$ and $\left( a ^2+ ab , b ^2+ ab ,- ab \right)$ are coplanar, where none of $a , b$ and c is zero, then
Answer(B) Since the given vectors are coplanar,
$\therefore\left|\begin{array}{ccc}-b c & b^2+b c & c^2+b c \\ a^2+a c & -a c & c^2+a c \\ a^2+a b & b^2+a b & -a b\end{array}\right|=0$
$\Rightarrow(a b+b c+c a)^3=0 \Rightarrow a b+b c+c a=0$
View full question & answer→MCQ 1272 Marks
The vectors $\lambda \hat{i}+\hat{j}+2 \hat{k}, \hat{i}+\lambda \hat{j}-\hat{k}$ and $2 \hat{ i }-\hat{ j }+\lambda \hat{ k }$ are coplanar if
- ✓
$\lambda=-2$
- B
$\lambda=0$
- C
$\lambda=2$
- D
$\lambda=1$
AnswerCorrect option: A. $\lambda=-2$
(A) Let $\overline{ a }, \overline{ b }$ and $\overline{ c }$ be the given vectors.
The given vectors are coplanar.
$\therefore\left|\begin{array}{ccc}\lambda & 1 & 2 \\ 1 & \lambda & -1 \\ 2 & -1 & \lambda\end{array}\right|=0$
$\begin{array}{l}\Rightarrow \lambda\left(\lambda^2-1\right)-(\lambda+2)+2(-1-2 \lambda)=0 \\ \Rightarrow \lambda^3-6 \lambda-4=0 \\ \Rightarrow(\lambda+2)\left(\lambda^2-2 \lambda-2\right)=0\end{array}$
$\Rightarrow \lambda=-2$ or $\lambda=\frac{2 \pm \sqrt{4+8}}{2}=1 \pm \sqrt{3}$
View full question & answer→MCQ 1282 Marks
If $\bar{a}, \bar{b}$ and $\bar{c}$ are non-coplanar vectors and the four points with position vectors $2 \bar{a}+3 \bar{b}-\bar{c}$, $\overline{ a }-2 \overline{b}+3 \overline{ c }, \quad 3 \overline{ a }+4 \overline{b}-2 \overline{ c }$ and $k \overline{ a }-6 \overline{b}+6 \overline{ c }$ are coplanar, then $k =$
Answer(B) Let $\bar{s}=2 \bar{a}+3 \bar{b}-\bar{c}, \bar{t}=\bar{a}-2 \bar{b}+3 \bar{c}$,
$\overline{ u }=3 \overline{ a }+4 \overline{b}-2 \overline{ c }, \overline{ v }= k \overline{ a }-6 \overline{b}+6 \overline{ c }$
$\begin{aligned} \therefore & \overline{ ST }=-\overline{ a }-5 \overline{b}+4 \overline{ c }, \overline{ SU }=\overline{ a }+\overline{ b }-\overline{ c } \\ & \overline{ SV }=( k -2) \overline{ a }-9 \overline{b}+7 \overline{ c }\end{aligned}$
Since the given points are coplanar, $[\overline{ ST }$ $\overline{ SU }$ $\overline{ SV }]=0$
$\Rightarrow\left|\begin{array}{ccc}-1 & -5 & 4 \\ 1 & 1 & -1 \\ k-2 & -9 & 7\end{array}\right|=0$
$\Rightarrow 2+5(7+ k -2)+4(-9- k +2)=0 \\ \Rightarrow 2+25+5 k -28-4 k =0 \\ \Rightarrow-1+ k =0 \\ \Rightarrow k =1$
View full question & answer→MCQ 1292 Marks
Given three arbitrary vectors $\bar{a}, \bar{b}, \bar{c}$, the vectors $\bar{\alpha}=5 \overline{ a }+6 \overline{b}+7 \overline{ c }, \beta=7 \overline{ a }-8 \overline{b}+9 \overline{ c }$, $\bar{\gamma}=3 \overline{ a }+20 \overline{b}+5 \overline{ c }$ are
Answer(B) $[\bar{\alpha} \bar{\beta} \bar{\gamma}]=\left|\begin{array}{ccc}5 & 6 & 7 \\ 7 & -8 & 9 \\ 3 & 20 & 5\end{array}\right|$
$=5(-40-180)-6(35-27)$ $+7(140+24)=0$
∴ the given vectors are coplanar.
View full question & answer→MCQ 1302 Marks
If the vectors $a \hat{i}+\hat{j}+\hat{k}, \hat{i}+b \hat{j}+\hat{k}$ and $\hat{ i }+\hat{ j }+ c \hat{ k }$ are coplanar $(a \neq b \neq c \neq 1)$, then the value of $a b c-(a+b+c)=$
Answer(B) Since $a \hat{i}+\hat{j}+\hat{k}, \quad \hat{i}+b \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+c \hat{k}$ are coplanar, $\left|\begin{array}{lll}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right|=0$
$\begin{array}{l}\Rightarrow a(b c-1)-1(c-1)+1(1-b)=0 \\ \Rightarrow a b c-a-b-c+2=0 \\ \Rightarrow a b c-(a+b+c)=-2\end{array}$
View full question & answer→MCQ 1312 Marks
If $\bar{x} \cdot \overline{ a }=0, \bar{x} \cdot \overline{b}=0$ and $\bar{x} \cdot \bar{c}=0$ for some non-zero vector $\bar{x}$, then the true statement is
- ✓
$[\overline{ a } \ \overline{ b } \ \overline{ c }]=0$
- B
[$\overline{a} \ \bar{b} \ \bar{c}$]$\neq 0$
- C
[$\overline{a} \ \bar{b} \ \bar{c}$] = 1
- D
AnswerCorrect option: A. $[\overline{ a } \ \overline{ b } \ \overline{ c }]=0$
(A) Since $\bar{x}$ is a non-zero vector, the given conditions will be satisfied, if either
i. at least one of the vectors $\bar{a}, \bar{b}, \bar{c}$ is zero or
ii. $\bar{x}$ is perpendicular to all the vectors $\overline{ a }, \overline{ b }, \overline{ c }$ In case (ii), $\overline{ a }, \overline{ b }, \overline{ c }$ are coplanar
$\Rightarrow[\overline{ a } \overline{ b } \overline{ c }]=0$
View full question & answer→MCQ 1322 Marks
If is is perpendicular to $\overline{ b }$ and $\overline{ c },|\overline{ a }|=2,|\overline{b}|=3$, $|\overline{c}|=4$. and the angle between $\bar{b}$ and $\bar{c}$ is $\frac{2 \pi}{3}$, then $\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$ is equal to
- A
$4 \sqrt{3}$
- B
$6 \sqrt{3}$
- ✓
$12 \sqrt{3}$
- D
$18 \sqrt{3}$
AnswerCorrect option: C. $12 \sqrt{3}$
(C) Let $\hat{ n }$ be the unit vector perpendicular to $\overline{ a }$ and $\overline{ b }$
$[\overline{ a } \overline{ b } \overline{ c }]=\overline{ a } \cdot(\overline{ b } \times \overline{ c })=\overline{ a } \cdot(|\overline{ b }||\overline{ c }| \sin \theta \hat{ n })$
$=\overline{ a } \cdot\left(3 \times 4 \sin \frac{2 \pi}{3} \cdot \hat{ n }\right)=\overline{ a } \cdot\left(12 \times \frac{\sqrt{3}}{2} \hat{ n }\right)$
$=6 \sqrt{3}|\overline{ a }||\hat{ n }| \cos 0=6 \sqrt{3} \times 2 \times 1=12 \sqrt{3}$
View full question & answer→MCQ 1332 Marks
Let $\overline{ a }=\hat{ i }-\hat{ k }, \overline{ b }=x \hat{ i }+\hat{ j }+(1-x) \hat{ k }$ and $\overline{ c }=y \hat{ i }+x \hat{ j }+(1+x-y) \hat{ k }$ then $[\overline{ a } \overline{ b } \overline{ c }]$ depends on:
- A
only $x$
- B
only $y$
- ✓
neither $x$ nor $y$
- D
both $x$ and $y$
AnswerCorrect option: C. neither $x$ nor $y$
(C) $\left[\begin{array}{ccc}\overline{ a } & \overline{ b } & \overline{ c }\end{array}\right]=\left|\begin{array}{ccc}1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y\end{array}\right|$
Applying $C _3 \Rightarrow C _3+ C _1$, we get
$[\overline{ a } \overline{ b } \overline{ c }]=\left|\begin{array}{ccc}1 & 0 & 0 \\ x & 1 & 1 \\ y & x & 1+x\end{array}\right|=1(1+x-x)=1$
View full question & answer→MCQ 1342 Marks
If the angle between the vectors $\bar{a}$ and $\bar{b}$ is $\frac{\pi}{3}$ and the area of the triangle with adjacent sides parallel to $\bar{a}$ and $\bar{b}$ is 3 , then $\bar{a} \cdot \bar{b}$ is
- A
$\sqrt{3}$
- ✓
$2 \sqrt{3}$
- C
$4 \sqrt{3}$
- D
$\frac{\sqrt{3}}{2}$
AnswerCorrect option: B. $2 \sqrt{3}$
(B) Given, $\frac{1}{2}|\bar{a} \times \bar{b}|=3$
$\Rightarrow|\bar{a} \times \bar{b}|=6$
$\Rightarrow|\bar{a}||\bar{b}| \sin \frac{\pi}{3}=6$
$\Rightarrow|\bar{a}||\bar{b}|=4 \sqrt{3}$
$\therefore \quad \bar{a} \cdot \bar{b}=|\bar{a}||\bar{b}| \cos \frac{\pi}{3}=4 \sqrt{3} \times \frac{1}{2}=2 \sqrt{3}$
View full question & answer→MCQ 1352 Marks
If $\bar{a}=\hat{i}+2 \hat{j}+2 \hat{k},|\bar{b}|=5$ and the angle between $\overline{ a }$ and $\overline{ b }$ is $\frac{\pi}{6}$, then the area of the triangle formed by these two vectors as two sides is
- ✓
$\frac{15}{4}$
- B
$\frac{15}{2}$
- C
$\frac{15 \sqrt{3}}{2}$
- D
$15$
AnswerCorrect option: A. $\frac{15}{4}$
(A) Area of triangle $=\frac{1}{2}|\overline{ a } \times \overline{ b }|$
$=\frac{1}{2}|\overline{ a }||\overline{ b }| \sin \theta$
$=\frac{1}{2} \times 3 \times 5 \times \sin \frac{\pi}{6}$
$=\frac{15}{4}$
View full question & answer→MCQ 1362 Marks
Area of $\Delta$ with vertices $(1,0,0),(0,1,0)$ and $(0,0,1)$ is
AnswerCorrect option: C. $\frac{\sqrt{3}}{2}$ sq. units
(C) For $\Delta$ with vertices $\overline{ a }, \overline{ b }$ and $\overline{ c }$, the area is given by $\frac{1}{2}|(\bar{b}-\bar{a}) \times(\bar{c}-\bar{a})|$
Here, $\overline{ a }=\hat{ i }, \overline{ b }=\hat{ j }, \overline{ c }=\hat{ k }$
$\therefore \quad \overline{ b }-\overline{ a }=(-\hat{ i }+\hat{ j })$ and $\overline{ c }-\overline{ a }=(-\hat{ i }+\hat{ k })$
$\therefore \quad$ area of $\Delta=\frac{1}{2}\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right|=\frac{1}{2}|\hat{ i }+\hat{ j }+\hat{ k }|$
$=\frac{\sqrt{3}}{2}$ sq. units
View full question & answer→MCQ 1372 Marks
Let $\overline{ a }, \overline{ b }, \overline{ c }$ be three vectors such that $\overline{ a } \neq 0$, and $\bar{a} \times \bar{b}=2 \bar{a} \times \bar{c},|\bar{a}|=|\overline{d}|=1,|\overline{b}|=4$ and $|\overline{ b } \times \overline{ c }|=\sqrt{15}$. If $\overline{ b }-2 \overline{ c }=\lambda \overline{ a }$, then $\lambda$ is equal to
AnswerCorrect option: B. $\pm 4$
(B) If angle between $\bar{b}$ and $\bar{c}$ is $\alpha$ and $|\overline{ b } \times \overline{ c }|=\sqrt{15}$
$\Rightarrow|\overline{ b }||\overline{ c }| \sin \alpha=\sqrt{15}$
$\Rightarrow \sin \alpha=\frac{\sqrt{15}}{4}$
$\Rightarrow \cos \alpha=\frac{1}{4}$
Now, $\overline{ b }-2 \overline{ c }=\lambda \overline{ a }$
$\Rightarrow|\overline{ b }-2 \overline{ c }|^2=\lambda^2|\overline{b}|^2$
$\begin{array}{l}\Rightarrow|\overline{ b }|^2+4|\overline{ c }|^2-4 \overline{b} \cdot \overline{ c }=\lambda^2|\overline{ a }|^2 \\ \Rightarrow 16+4-4\{|\overline{b}||\overline{ c }| \cos \alpha\}=\lambda^2\end{array}$
$\Rightarrow 16+4-4 \times 4 \times 1 \times \frac{1}{4}=\lambda^2$
$\begin{array}{l}\Rightarrow \lambda^2=16 \\ \Rightarrow \lambda= \pm 4\end{array}$
View full question & answer→MCQ 1382 Marks
Let $\bar{a}, \bar{b}$ and $\bar{c}$ be three unit vectors such that $\bar{a}+\bar{b}+\bar{c}=\overline{0}$. If $\lambda=\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}$ and $\overline{ d }=\overline{ a } \times \overline{ b }+\overline{ b } \times \overline{ c }+\overline{ c } \times \overline{ a }$, then the ordered pair, $(\lambda, \bar{d})$ is equal to
- A
$\left(-\frac{3}{2}, 3 \bar{c} \times \bar{b}\right)$
- ✓
$\left(-\frac{3}{2}, 3 \overline{ a } \times \overline{ b }\right)$
- C
$\left(\frac{3}{2}, 3 \bar{b} \times \bar{c}\right)$
- D
$\left(\frac{3}{2}, 3 \overline{ a } \times \overline{ c }\right)$
AnswerCorrect option: B. $\left(-\frac{3}{2}, 3 \overline{ a } \times \overline{ b }\right)$
(B) $\bar{a}+\bar{b}+\bar{c}=\overline{0}$
$\begin{array}{l}|\overline{ a }+\overline{ b }+\overline{ c }|^2=0 \\ |\overline{ a }|^2+|\overline{ b }|^2+|\overline{ c }|^2+2(\overline{ a } \cdot \overline{ b }+\overline{ b } \cdot \overline{ c }+\overline{ c } \cdot \overline{ a })=0\end{array}$
$3+2(\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a})=0$
$\overline{ a } \cdot \overline{ b }+\overline{ b } \cdot \overline{ c }+\overline{ c } \cdot \overline{ a }=-\frac{3}{2}$
$=-\frac{3}{2}$
$\overline{ d }=\overline{ a } \times \overline{ b }+\overline{ b } \times \overline{ c }+\overline{ c } \times \overline{ a }$
$\begin{array}{l}=\bar{a} \times \bar{b}+\bar{b} \times(-\bar{a}-\bar{b})+(-\bar{a}-\bar{b}) \times \bar{a} \\ =\bar{a} \times \bar{b}-\bar{b} \times \bar{a}-\bar{b} \times \bar{a}=3(\bar{a} \times \bar{b})\end{array}$
View full question & answer→MCQ 1392 Marks
If $\bar{a}, \bar{b}, \bar{c}$ are three vectors such that $|\overline{ a }+\overline{ b }+\overline{ c }|=1, \quad \overline{ c }=\lambda$ $(\bar{a} \times \bar{b})$ and $|\bar{a}|=\frac{1}{\sqrt{2}}$, $|\overline{b}|=\frac{1}{\sqrt{3}},|\overline{c}|=\frac{1}{\sqrt{6}}$, then the angle between $\overline{a}$ and $\bar{b}$ is
- A
$\frac{\pi}{6}$
- B
$\frac{\pi}{4}$
- C
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{2}$
AnswerCorrect option: D. $\frac{\pi}{2}$
(D) Let $\theta$ be the angle between $\overline{ a }$ and $\overline{ b }$.
Since $\overline{ c }=\lambda(\overline{ a } \times \overline{ b })$
$\begin{array}{l}\Rightarrow \overline{ c } \perp \overline{ a }, \overline{ c } \perp \overline{ b } \\ \Rightarrow \overline{ c } \cdot \overline{ a }=\overline{ c } \cdot \overline{ b }=0\end{array}$
Now,
$|\overline{ a }+\overline{ b }+\overline{ c }|=1$
$\begin{array}{l}\Rightarrow|\overline{ a }+\overline{ b }+\overline{ c }|^2=1 \\ \Rightarrow|\overline{ a }|^2+|\overline{ b }|^2+|\overline{ c }|^2+2(\overline{ a } \cdot \overline{ b }+\overline{ b } \cdot \overline{ c }+\overline{ c } \cdot \overline{ a })=1 \\ \Rightarrow \frac{1}{2}+\frac{1}{3}+\frac{1}{6}+2\{|\overline{ a }||\overline{ b }| \cos \theta\}=1\end{array}$
$\Rightarrow \cos \theta=0$
$\Rightarrow \theta=\frac{\pi}{2}$
View full question & answer→MCQ 1402 Marks
If $\bar{u}$ and $\bar{v}$ are unit vectors and $\theta$ is the acute angle between them, then $2 \overline{ u } \times 3 \overline{ v }$ is a unit vector for
- A
no value of $\theta$
- ✓
exactly one value of $\theta$
- C
exactly two value of $\theta$
- D
more than two value of $\theta$
AnswerCorrect option: B. exactly one value of $\theta$
(B) Since $|\overline{ u }|=|\overline{ v }|=1$ and $\theta$ is the acute angle between $\overline{ u }$ and $\overline{ v }$.
$\therefore \quad|\bar{u} \times \bar{v}|=\sin \theta$ ...(i)
Now, $2 \overline{ u } \times 3 \overline{ v }$ will be a unit vector, if $|2 \overline{ u } \times 3 \overline{ v }|=1$
$\Rightarrow 6|\overline{ u } \times \overline{ v }|=1$
$\Rightarrow 6 \sin \theta=1$ ...[From (i)]
$\Rightarrow \sin \theta=\frac{1}{6}$
As $\theta$ is an acute angle. So, there is only one value of $\theta$ for which $2 \overline{ u } \times 3 \overline{ v }$ is a unit vector.
View full question & answer→MCQ 1412 Marks
If $\bar{a}, \bar{b}$ and $\bar{c}$ are unit vectors such that $\overline{ a }+\overline{ b }+\overline{ c }=\overline{0}$ and angle between $\overline{ a }$ and $\overline{ b }$ is $\frac{\pi}{3}$, then $|\overline{ a } \times \overline{ b }|+|\overline{ b } \times \overline{ c }|+|\overline{ c } \times \overline{ a }|=$
- A
$\frac{3}{2}$
- B
$0$
- ✓
$\frac{3 \sqrt{3}}{2}$
- D
AnswerCorrect option: C. $\frac{3 \sqrt{3}}{2}$
(C) Since $\bar{a}+\bar{b}+\bar{c}=0$
$\therefore \quad \overline{ a } \times \overline{ b }=\overline{ b } \times \overline{ c }=\overline{ c } \times \overline{ a }$
$|\overline{ a } \times \overline{ b }|+|\overline{ b } \times \overline{ c }|+|\overline{ c } \times \overline{ a }|=3|\overline{ a } \times \overline{ b }|$
$\begin{array}{l}=3|\overline{ a }||\overline{ b }| \sin \frac{\pi}{3} \\ =\frac{3 \sqrt{3}}{2}\end{array}$
View full question & answer→MCQ 1422 Marks
Vectors $\bar{a}$ and $\bar{b}$ are inclined at an angle $\theta=60^{\circ}$. If $|\overline{ a }|=1,|\overline{b}|=2$, then $[(\bar{a}+3 \bar{b}) \times(3 \bar{a}-\bar{b})]^2$ is equal to
Answer(D) $[(\bar{a}+3 \bar{b}) \times(3 \bar{a}-\bar{b})]^2$
$\begin{array}{l}=[10(\overline{b} \times \overline{ a })]^2 \\ =100|\overline{b} \times \overline{ a }|^2\end{array}$
$=100\left\{|\overline{ a }|^2|\overline{b}|^2-(\overline{ a } \cdot \overline{ b })^2\right\}$
$\begin{array}{l}=100(4-1) \quad \ldots .\left[\because \bar{a} \cdot \bar{b}=2 \cos 60^{\circ}=1\right] \\ =300\end{array}$
View full question & answer→MCQ 1432 Marks
If $|\overline{ a }|=16,|\overline{b}|=4$, then $\sqrt{|\overline{ a } \times \overline{ b }|^2+|\overline{ a } \cdot \overline{ b }|^2}=$
Answer(B) $\sqrt{|\overline{ a } \times \overline{ b }|^2+|\overline{ a } \cdot \overline{ b }|^2}=\sqrt{|\overline{ a }|^2|\overline{b}|^2}$
If $\bar{a}, \bar{b}$ are two vectors, then
$(\overline{ a } \times \overline{ b })^2=\left|\begin{array}{ll}\overline{ a } \cdot \overline{ a } & \overline{ a } \cdot \overline{ b } \\ \overline{ a } \cdot \overline{ b } & \overline{ b } \cdot \overline{ b }\end{array}\right|$ or
$|\overline{ a } \times \overline{ b }|^2+|\overline{ a } \cdot \overline{ b }|^2=|\overline{ a }|^2|\overline{b}|^2$ where $0<\theta<\frac{\pi^{ c }}{2}$
$=|\overline{ a }||\overline{ b }|=(16)(4)$
$=64$
View full question & answer→MCQ 1442 Marks
If $\bar{a}$ and $\bar{b}$ are two vectors such that $\bar{a} \cdot \bar{b}=0$ and $\bar{a} \times \bar{b}=0$, then
AnswerCorrect option: C. Either a or b is a null vector
(C) $\overline{ a } \cdot \overline{ b }=0$
$\begin{array}{l}\Rightarrow \overline{ a } \perp \overline{ b } \\ \Rightarrow \overline{ a }=0 \text { or } \overline{ b }=0\end{array}$
and $\overline{ a } \times \overline{ b }=0$
$\begin{array}{l}\Rightarrow \overline{ a } \| \overline{ b } \\ \Rightarrow \overline{ a }=0 \text { or } \overline{ b }=0\end{array}$
Hence, either $\bar{a}$ or $\bar{b}$ is a null vector.
View full question & answer→MCQ 1452 Marks
If $\bar{a}+\bar{b}+\bar{c}=\overline{0}$, then which relation is correct?
- A
$\overline{ a }=\overline{ b }=\overline{ c }=0$
- B
$\overline{ a } \cdot \overline{ b }=\overline{ b } \cdot \overline{ c }=\overline{ c } \cdot \overline{ a }$
- ✓
$\overline{ a } \times \overline{ b }=\overline{ b } \times \overline{ c }=\overline{ c } \times \overline{ a }$
- D
AnswerCorrect option: C. $\overline{ a } \times \overline{ b }=\overline{ b } \times \overline{ c }=\overline{ c } \times \overline{ a }$
(C) Since $\bar{a}+\bar{b}+\bar{c}=\overline{0}$
$\begin{array}{l}\Rightarrow \overline{ a } \times(\overline{ a }+\overline{ b }+\overline{ c })=0 \\ \Rightarrow \overline{ a } \times \overline{ a }+\overline{ a } \times \overline{ b }+\overline{ a } \times \overline{ c }=0\end{array}$
$\Rightarrow \overline{ a } \times \overline{ b }=-\overline{ a } \times \overline{ c }=\overline{ c } \times \overline{ a }$ ...(i)
Similarly, $\overline{ b } \times(\overline{ a }+\overline{ b }+\overline{ c })=0$
$\Rightarrow \overline{ a } \times \overline{ b }=\overline{ b } \times \overline{ c }$ ...(ii)
By (i) and (ii), we get
$\overline{a} \times \overline{b}=\overline{b} \times \overline{c}=\overline{c} \times \overline{a}$
View full question & answer→MCQ 1462 Marks
If $\bar{a}=\hat{i}+\hat{j}+\hat{k}, \bar{a} \cdot \bar{b}=1$ and $\bar{a} \times \bar{b}=\hat{j}-\hat{k}$, then $\bar{b}=$
AnswerCorrect option: A. $\hat{i}$
(A) Let $\overline{ b }= b _1 \hat{ i }+ b _2 \hat{ j }+ b _3 \hat{ k }$
Now, $\hat{j}-\hat{k}=\bar{a} \times \bar{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ b_1 & b_2 & b_3\end{array}\right|$
$\begin{array}{l}\Rightarrow b _3- b _2=0, b_1- b _3=1, b_2- b _1=-1 \\ \Rightarrow b_3= b _2, b_1= b _2+1\end{array}$
Also, $\overline{ a } \cdot \overline{ b }=1$
$\begin{array}{l}\Rightarrow b_1+b_2+b_3=1 \\ \Rightarrow 3 b_2+1=1 \\ \Rightarrow b_2=0 \\ \Rightarrow b_1=1, b_3=0\end{array}$
Thus, $\overline{ b }=\hat{ i }$
View full question & answer→MCQ 1472 Marks
If $\bar{a}=x \hat{i}+y \hat{j}+z \hat{k}$, then $(\bar{a} \times \hat{i}) \cdot(\bar{i}+\hat{j})+(\bar{a} \times \hat{j}) \cdot(\hat{j}+\hat{k})+(\bar{a} \times \hat{k}) \cdot(\hat{k}+\hat{i})=$
- A
$x-y+z$
- ✓
$x+y+z$
- C
$x+y-z$
- D
$-x+y+z$
AnswerCorrect option: B. $x+y+z$
(B) $(\overline{ a } \times \hat{ i }) \cdot(\hat{ i }+\hat{ j })+(\overline{ a } \times \hat{ j }) \cdot(\hat{ j }+\hat{ k })+(\overline{ a } \times \hat{ k }) \cdot(\hat{ k }+\hat{ i })$
$=(-y \hat{ k }+z \hat{ j }) \cdot(\hat{ i }+\hat{ j })+(x \hat{ k }-z \hat{ i }) \cdot(\hat{ j }+\hat{ k })$ $+(-x \hat{ j }+y \hat{ i }) \cdot(\hat{ k }+\hat{ i })$
$= z +x+y$
View full question & answer→MCQ 1482 Marks
If $\bar{a}$ is a unit vector, then $|\overline{ a } \times \hat{ i }|^2+|\overline{ a } \times \hat{ j }|^2+|\overline{ a } \times \hat{ k }|^2=$
Answer(A) $|\overline{ a } \times \hat{ i }|^2+|\overline{ a } \times \hat{ j }|^2+|\overline{ a } \times \hat{ k }|^2=2|\overline{ a }|^2=2$
View full question & answer→MCQ 1492 Marks
For any vector $\bar{x}$ the value of $(\bar{x} \times \hat{ i })^2+(\bar{x} \times \hat{ j })^2+(\bar{x} \times \hat{ k })^2$ is equal to
- A
$|\bar{x}|^2$
- ✓
$2|\bar{x}|^2$
- C
$3|\bar{x}|^2$
- D
$4|\bar{x}|^2$
AnswerCorrect option: B. $2|\bar{x}|^2$
(B) $|x \times \hat{ i }|^2=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ x_1 & x_2 & x_3 \\ 1 & 0 & 0\end{array}\right| \ldots\left[\because x=x_{ i } \hat{ i }+x_2 \hat{ j }+x_3 \hat{ k }\right]$
$=\left|x_3 \hat{ j }-x_2 \hat{ k }\right|^2=x_3^2+x_2^2$
Similarly, $|\bar{x} \times \hat{j}|^2=x x_1^2+x_3^2$ and $|\bar{x} \times \hat{ k }|^2=x_1^2+x_2^2$Hence, the required result is
$=2\left(x_1^2+x_2^2+x_3^2\right)=2|\bar{x}|^2$
View full question & answer→MCQ 1502 Marks
$\hat{ i } \cdot(\hat{ j } \times \hat{ k })+\hat{ j } \cdot(\hat{ i } \times \hat{ k })+\hat{ k } \cdot(\hat{ i } \times \hat{ j })$ is equal to
Answer(D) $\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})$
$ =\hat{i} \cdot \hat{i}+\hat{j} \cdot \hat{j}+\hat{k} \cdot \hat{k}=3$
$=1-1+1=1$
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