2 Marks Questions

Question 12 Marks

Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{b}=2 \hat{i}-7 \hat{j}+\hat{k}$.

Answer

According to question, the adjacent sides of parallelogram,
and
$\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$
$ \begin{aligned} \therefore \vec{a} \times \vec{b} & =\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\
2 & -7 & 1 \end{array}\right| \\ & =(-1+21) \hat{i}-(1-6) \hat{j}+(-7+2) \hat{k} \\ & =20 \hat{i}+5 \hat{j}-5 \hat{k} \end{aligned} $
$\therefore$ Area of parallelogram
$=|\vec{a} \times \vec{b}|=|20 \hat{i}+5 \hat{j}-5 \hat{k}|$
$=\sqrt{20^2+5^2+(-5)^2}$
$=\sqrt{400+25+25}$
$=\sqrt{400}$
$=15 \sqrt{2} \text { sq. unit. }$

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Question 22 Marks

Evaluate the product $(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})$

Answer

According to question,
$(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})$
$\Rightarrow\binom{3}{3} \cdot(2 \vec{b})+\binom{3}{3} \cdot(7 \vec{b})+(-5 \vec{b}) \cdot(2 \vec{a})+(-5 \vec{b}) \cdot(7 \vec{b})$
$=6 \vec{a} \cdot \vec{a}+21 \vec{a} \cdot \vec{b}-10 \vec{b} \cdot \vec{a}-35 \vec{b} \cdot \vec{b}$
$=6|\vec{a}|^2+21(\vec{a} \cdot \vec{b})-10(\vec{b} \cdot \vec{a})-35|\vec{b}|^2$
$\because \vec{a} \cdot \vec{a}=|\vec{a}|^2, \vec{b} \cdot \vec{b}=|\vec{b}|^2, \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}$
So, $(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})$
$=6|\vec{a}|^2+21(\vec{a} \cdot \vec{b})_{-10} \vec{a} \cdot \vec{b}-35|\vec{b}|^2$
$=6|\vec{a}|^2+11 \vec{a} \cdot \vec{b}-35|\vec{b}|^2$

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Question 32 Marks

If $|\vec{a}|=10,|\vec{b}|=2$ and $\vec{a} \cdot \vec{b}=12$, then find the value of $|\vec{a} \times \vec{b}|$.

Answer

We know that
$
|\vec{a} \times \vec{b}|=a b \sin \theta=|\vec{a}||\vec{b}| \sin \theta
$
On putting the values $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$
$
12=10 \times 2 \times \cos \theta
$
Then
$
\begin{aligned}
\cos \theta & =\frac{12}{20}=\frac{3}{5} \\
\sin \theta & =\sqrt{1-\cos ^2 \theta}=\sqrt{1-\frac{9}{25}} \\
\sin \theta & =\sqrt{\frac{16}{25}}=\frac{4}{5} \\
|\vec{a} \times \vec{b}| & =|\vec{a}||\vec{b}| \sin \theta
\end{aligned}
$
Putting the values
$
=(10) \times(2) \times \frac{4}{5}=16
$
Hence $|\vec{a} \times \vec{b}|=16$

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Question 42 Marks

If two vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}|=2,|\vec{b}|=3$ and $\vec{a} \cdot \vec{b}=4$, then find $|\vec{a}-\vec{b}|$.

Answer

We have
$
\begin{aligned}
|\vec{a}-\vec{b}|^2 & =(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b}) \\
& =\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b} \\
& =|\vec{a}|^2-2(\vec{a} \cdot \vec{b})+|\vec{b}|^2 \\
& =(2)^2-2(4)+(3)^2 \\
& =4-8+9 \\
& =5
\end{aligned}
$
Therefore $|\vec{a}-\vec{b}|=\sqrt{5}$

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Question 52 Marks

Find the unit vector along the sum of the vectors $\vec{a}=2 \hat{i}+2 \hat{j}-5 \hat{k}$ and $\vec{b}=2 \hat{i}+2 \hat{j}+\hat{k}$.

Answer

$\vec{a}+\vec{b} =2 \hat{i}+2 \hat{j}-5 \hat{k}+2 \hat{i}+\hat{j}+3 \hat{k}$
$\vec{a}+\vec{b} =4 \hat{i}+3 \hat{j}-2 \hat{k}$
$|\vec{a}+\vec{b}| =\sqrt{(4)^2+(3)^2+(-2)^2}$
$ =\sqrt{16+9+4}=\sqrt{29}$
Hence unit vector along $\vec{a}+\vec{b}$
$=\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}$
$=\frac{4 \hat{i}+3 \hat{j}-2 \hat{k}}{\sqrt{29}}$
$=\frac{4}{\sqrt{29}} \hat{i}+\frac{3}{\sqrt{29}} \hat{j}-\frac{2}{\sqrt{29}} \hat{k}$

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Question 62 Marks

If $\overrightarrow{ a }=2 \hat{ i }+2 \hat{ j }+3 \hat{ k }, \overrightarrow{ b }=-\hat{ i }+2 \hat{ j }+\hat{ k }$ and $\overrightarrow{ c }=3 \hat{ i }+\hat{ j }$ be such that $\vec{a}+\lambda b$ is perpendicular to vector $\vec{c}$, then find the value of $\lambda$.

Answer

According to the question and
$ \vec{a} =2 \hat{i}+2 \hat{j}+3 \hat{k}$
$ \vec{b} =-\hat{i}+2 \hat{j}+\hat{k}$
$\therefore \vec{a}+\lambda \vec{b} =(2 \hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})$
$ =(2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \hat{k} $
and given that $\vec{c}=3 \hat{i}+\hat{j}$
$ \because \vec{a}+\lambda \vec{b} \perp \vec{c} ($ given $)$
$\therefore(\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0$
$\Rightarrow[(2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \hat{k}] \cdot[3 \hat{i}+\hat{j}]=0$
$\Rightarrow 3(2-\lambda)+(2+2 \lambda) \cdot 1+(3+\lambda) \cdot 0=0$
$\Rightarrow 6-3 \lambda+2+2 \lambda=0$
$\Rightarrow 8-\lambda=0$
$\Rightarrow \lambda=8 $

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Question 72 Marks

Prove that $\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b})=0$

Answer

Using the distributive law of vector product,
$
\begin{aligned}
& LHS=\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b}) \\
= & (\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})+(\vec{b} \times \vec{c})+(\vec{b} \times \vec{a})+(\vec{c} \times \vec{a})+(\vec{c} \times \vec{b}) \\
= & (\vec{a} \times \vec{b})-(\vec{c} \times \vec{a})+(\vec{b} \times \vec{c})-(\vec{a} \times \vec{b})+(\vec{c} \times \vec{a})-(\vec{b} \times \vec{c}) \\
= & 0=\operatorname{RHS} \quad \text { Hence proved. }
\end{aligned}
$

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Question 82 Marks

Prove that the vectors $\vec{a}-2 \vec{b}+3 \vec{c}, 2 \vec{a}+3 \vec{b}-4 \vec{c}$ and $-7 \vec{b}+10 \vec{c}$ are collinear.

Answer

Let the position vectors of points $A , B , C$ with respect to O be $\vec{a}-2 \vec{b}+3 \vec{c}, 2 \vec{a}+3 \vec{b}-4 \vec{c},-7 \vec{b}+10 \vec{c}$ respectively, i.e.,
$
\overrightarrow{OA}=\vec{a}-2 \vec{b}+3 \vec{c}, \overrightarrow{OB}=2 \vec{a}+3 \vec{b}-4 \vec{c}, \overrightarrow{OC}=-7 b+10 \vec{c}
$
Then $\overrightarrow{ AB }=\overrightarrow{ OB }-\overrightarrow{ OA }=(2 \vec{a}+3 \vec{b}-4 \vec{c})-(\vec{a}-2 \vec{b}+3 \vec{c})$
$
=\vec{a}+5 \vec{b}-7 \vec{c}
$
and
$
\begin{aligned}
\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA} =(-7 \vec{b}+10 \vec{c})-(\vec{a}-2 \vec{b}+3 \vec{c}) \\
=-\vec{a}-5 \vec{b}+7 \vec{c} \\
=-(\vec{a}+5 \vec{b}-7 \vec{c})=-\overrightarrow{AB}
\end{aligned}
$
$\therefore \quad \overrightarrow{ AC }=-\overrightarrow{ AB } \Rightarrow \overrightarrow{ AC }=m \overrightarrow{ AB } \quad$ [where $m=-1$ ]
It shows that the vectors $\overrightarrow{A C}$ and $\overrightarrow{A B}$ and parallel which is possible only when the all three points $A, B, C$ lie in one line, since the initial point A of both of these vectors is common. Hence proved.

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Question 92 Marks

Write a vector parallel to the vector $2 \hat{i}-\hat{j}$ whose magnitude is $5$ units.

Answer

Let the vector parallel to the vector $2 \hat{i}-\hat{j}$ be $a_1 \hat{i}+a_2 \hat{j}$. Given that
$\left|a_1 \hat{i}+a_2 \hat{j}\right| =5$
$\because |2 \hat{i}-\hat{j}| =\sqrt{(2)^2+(-1)^2}=\sqrt{5}$
$\sqrt{a_1^2+a_2^2} =5$
$a_1^2+a_2^2 =25$
For being parallel :
$\frac{a_1}{2} =\frac{a_2}{-1}=k$
$a_1 =2 k$
$a_2 =-k$
Putting the values in equation $(1),$
$(2 k)^2+(-k)^2=25$
$\Rightarrow 4 k^2+k^2=25$
$5 k^2=25$
$k=\sqrt{5}$
$\therefore a_1=2 \sqrt{5}$ and $a_2=-\sqrt{5}$
Hence the parallel vector will be $2 \sqrt{5} \hat{i}-\sqrt{5} \hat{j}$.

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