Maths MCQ

$\begin{aligned} & \text {(d) : We have, } \frac{(p \hat{i}-\hat{j}+2 \hat{k}) \cdot(2 \hat{i}-p \hat{j}-\hat{k})}{\sqrt{p^2+1+4} \sqrt{4+p^2+1}}=\cos \frac{\pi}{3} \\ & \Rightarrow \frac{2 p+p-2}{\sqrt{p^2+5} \sqrt{p^2+5}}=\cos \frac{\pi}{3} \Rightarrow \frac{3 p-2}{p^2+5}=\frac{1}{2} \\ & \Rightarrow 6 p-4=p^2+5 \Rightarrow p^2-6 p+9=0 \Rightarrow p=3\end{aligned}$

(c) : Let equation of planes be $P_1=x-c y-b z$,
$\begin{aligned}
& P_2=c x-y+a z=0, P_3=b x+a y-z=0 . \\
& \therefore\left|\begin{array}{ccc}
1 & -c & -b \\
c & -1 & a \\
b & a & -1
\end{array}\right|=0 \\
& \Rightarrow 1\left(1-a^2\right)+c(-c-a b)-b(a c+b)=0 \\
& \Rightarrow 1-a^2-c^2-a b c-a b c-b^2=0 \\
& \Rightarrow 1-2 a b c-a^2-b^2-c^2=0 \Rightarrow a^2+b^2+c^2=1-2 a b c
\end{aligned}$

(a) : We have, $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z-0}{1}$ intersect each other.
$\begin{aligned}
& \Rightarrow\left|\begin{array}{ccc}
3-1 & k+1 & 0-1 \\
2 & 3 & 4 \\
1 & 2 & 1
\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}
2 & k+1 & -1 \\
2 & 3 & 4 \\
1 & 2 & 1
\end{array}\right|=0 \\
& \Rightarrow 2(3-8)-(k+1)(2-4)-1(4-3)=0 \Rightarrow k=\frac{9}{2}
\end{aligned}
$

(c): The required line is perpendicular to the lines which are parallel to the vectors $\vec{b}_1=2 \hat{i}-2 \hat{j}+\hat{k}$ and $\vec{b}_2=\hat{i}-2 \hat{j}+2 \hat{k}$. So, it is parallel to the vector $\vec{b}=\vec{b}_1 \times \vec{b}_2$ Now, $\vec{b}=\vec{b}_1 \times \vec{b}_2=(-4+2) \hat{i}-(4-1) \hat{j}+(-4+2) \hat{k}$
$
=-2 \hat{i}-3 \hat{j}-2 \hat{k}
$
Thus, the required line passes through the point $(3,-1,2)$ and is parallel to the vector $\vec{b}=-2 \hat{i}-3 \hat{j}-2 \hat{k}$
$\therefore \quad$ Equation of line is $\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-2}{2}$

(c) : Direction ratios of $\vec{n}_1$ are $m,-1,2$
Direction ratios of $\vec{n}_2$ are $2,-m,-1$
Now,$\begin{aligned} & \cos \theta=\left|\frac{\vec{n}_1 \cdot \vec{n}_2}{\left|\vec{n}_1\right|\left|\vec{n}_2\right|}\right| \Rightarrow \frac{1}{2}=\left|\frac{2 m+m-2}{\sqrt{m^2+5} \sqrt{m^2+5}}\right| \quad\left(\because \theta=\frac{\pi}{3}\right) \\
& \Rightarrow \frac{1}{2}=\left|\frac{3 m-2}{m^2+5}\right| \Rightarrow m^2+5= \pm(6 m-4) \\
& \Rightarrow m^2+5=6 m-4 \text { or } m^2+5=-6 m+4 \\
& \Rightarrow m^2-6 m+9=0 \text { or } m^2+6 m+1=0 \\
& \Rightarrow(m-3)^2=0 \Rightarrow m=3 \text { or } m=-3 \pm 2 \sqrt{2}
\end{aligned}
$

(a): The equation of plane is $\vec{r} \cdot(3 \hat{i}+2 \hat{j}+6 \hat{k})=13$ i.e., $3 x+2 y+6 z-13=0$
Now, the given point is $(2,3, \lambda) \equiv\left(x_1, y_1, z_1\right)$
Distance of the plane from the point $=5$ units
$\begin{aligned}
& \therefore \quad 5=\left|\frac{3(2)+2(3)+6 \lambda-13}{\sqrt{9+4+36}}\right| \Rightarrow 5=\left|\frac{6 \lambda-1}{7}\right| \\
& \Rightarrow \quad 6 \lambda-1= \pm 35 \Rightarrow 6 \lambda=36,6 \lambda=-34 \\
& \Rightarrow \quad \lambda=6, \lambda=-\frac{17}{3}
\end{aligned}$

 (a) : The equation of plane passing through $A(\vec{a})$ and $\perp$ to $\vec{n}$ is $\vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n}$ ...(i)
Let $\alpha$ be the equal acute angle that the normal makes with coordinate axes.
So, $l=m=n=\cos \alpha$
Now, $l^2+m^2+n^2=1$
$
\Rightarrow \cos ^2 \alpha+\cos ^2 \alpha+\cos ^2 \alpha=1 \Rightarrow \cos \alpha=\frac{1}{\sqrt{3}}
$
So, $l=m=n=\frac{1}{\sqrt{3}}$
Hence, dr's of $\vec{n}$ are $< 1,1,1>$
Now, $\vec{a}=-\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{n}=\hat{i}+\hat{j}+\hat{k}$
So, from (i), we get
$
\begin{aligned}
\quad \vec{r} \cdot(\hat{i}+\hat{j}+\hat{k}) & =(-\hat{i}+\hat{j}+2 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k}) \\
\Rightarrow \quad & \vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=-1+1+2=2
\end{aligned}
$

(b) : Equation of line passing through $\left(x_1, y_1 z_1\right)$
and having d.c.s. $l, m, n$ is $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$
Here $\left(x_1, y_1, z_1\right) \equiv(-3,2,-5)$
Also line is equally inclined to co-ordinate axes.
$\therefore \quad l=\frac{-1}{\sqrt{3}}, m=\frac{1}{\sqrt{3}}, n=\frac{-1}{\sqrt{3}}$
$\therefore$ Equation of line is $\frac{x+3}{\frac{-1}{\sqrt{3}}}=\frac{y-2}{\frac{1}{\sqrt{3}}}=\frac{z+5}{\frac{-1}{\sqrt{3}}}$
or $\frac{x+3}{-1}=\frac{y-2}{1}=\frac{z+5}{-1}$

(b) : The equation of the given lines are
$\frac{x-1} {2}=\frac{y+1}{2}=\frac{z-1}{4}=r \text { (say) ...(i) }$
and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=t$ (say) ...(ii)
Any point on (i) is $(2 r+1,2 r-1,4 r+1)$
Any point on (ii) is $(t+3,2 t+k, t)$
(i) and (ii) will intersect if
$\begin{aligned}
& 2 r+1=t+3,2 r-1=2 t+k, 4 r+1=t \\
& \Rightarrow 2 r+1=4 r+1+3 \Rightarrow r=-\frac{3}{2}
\end{aligned}$
So, $t=4\left(-\frac{3}{2}\right)+1=-5$
Thus, required point of intersection is $(-2,-4,-5)$.

(b) : Let the angle between line $\vec{r}=\vec{a}+\lambda \vec{b}$ and the plane $\vec{r} \cdot \vec{n}=0$ is $\theta$, then
$\begin{aligned}
& \sin \theta=\frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|}=\left|\frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{1^2+1^2+1^2}} \cdot \frac{(2 \hat{i}-\hat{j}+\hat{k})}{\sqrt{2^2+(-1)^2+1^2}}\right| \\
& =\left|\frac{2-1+1}{\sqrt{3} \sqrt{6}}\right|=\left|\frac{2}{3 \sqrt{2}}\right|=\frac{\sqrt{2}}{3} \quad \therefore \theta=\sin ^{-1}\left(\frac{\sqrt{2}}{3}\right)
\end{aligned}
$

$\begin{aligned} & \text {(a) }: A \equiv(0, b, c) \text { and } B \equiv(a, 0, c) \\ & \therefore \quad \overrightarrow{O A}=b \hat{j}+c \hat{k} \text { and } \overrightarrow{O B}=a \hat{i}+c \hat{k} \\ & \therefore \quad \vec{n}=\overrightarrow{O A} \times \overrightarrow{O B}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 0 & b & c \\ a & 0 & c\end{array}\right| \\ & =\hat{i}(b c)-\hat{j}(-a c)+\hat{k}(-a b)=(b c) \hat{i}+(a c) \hat{j}-(a b) \hat{k} \\ & \therefore \quad Eq ^{ n } \text { of plane is } \\ & \vec{r} \cdot(b c \hat{i}+a c \hat{j}-a b \hat{k})=0 \\ & \Rightarrow(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(b c \hat{i}+a c \hat{j}-a b \hat{k})=0 \\ & \Rightarrow \quad x b c+y a c-z a b=0 \Rightarrow \frac{x}{a}+\frac{y}{b}-\frac{z}{c}=0\end{aligned}$

(b) : Let $l, m, n$, be direction cosines of the line.
$\therefore \quad-l+2 m+2 n=0$ and $2 m+n=0$
$\Rightarrow m=-\frac{n}{2} \therefore \frac{m}{-1}=\frac{n}{2}$ ...(i)
Also $-l+2 m+2 n=0 \Rightarrow-l-n+2 n=0$
$\Rightarrow-l+n=0 \Rightarrow l=n \Rightarrow \frac{l}{2}=\frac{n}{2}$ ...(ii)
From (i) and (ii)
$\frac{l}{2}=\frac{m}{-1}=\frac{n}{2}$
Hence, direction ratios are $(2,-1,2)$

(a) : Since, equation of line is
$\frac{x+2}{2}=\frac{2 y-5}{3}, z=-1 \Rightarrow \frac{x+2}{2}=\frac{y-5 / 2}{3 / 2}=\frac{z+1}{0}$
$\therefore \quad$ Direction cosines of the line are
$\frac{2}{\sqrt{(2)^2+\left(\frac{3}{2}\right)^2+(0)^2}}, \frac{3 / 2}{\sqrt{(2)^2+\left(\frac{3}{2}\right)^2+(0)^2}}$
$=\frac{2}{5 / 2}, \frac{3 / 2}{5 / 2}, \frac{0}{5 / 2}=\left(\frac{4}{5}, \frac{3}{5}, 0\right) \quad \frac{0}{\sqrt{(2)^2+\left(\frac{3}{2}\right)^2+(0)^2}}$

 (b) : Given points are $P(2,3,4), Q(-1,-2,1)$ and $R(5,8,7)$.
Direction ratios of $P Q$ are $(-1-2,-2-3,1-4)$, i.e., ( -3 , $-5,-3$)
Direction ratios of $Q R$ are $(5+1,8+2,7-1)$, i.e., (6, 10,6 )
As $\frac{-3}{6}=\frac{-5}{10}=\frac{-3}{6}$, therefore the direction ratios of $P Q$ and $Q R$ are proportional.
Hence, $P Q$ and $Q R$ are parallel but they have a point $Q$ in common. Therefore $P Q$ and $Q R$ are along the same line i.e., $P, Q$ and $R$ are collinear.

$
\begin{aligned}
& \text { (b) : Here } a_1=2, b_1=1, c_1=-3 \\
& a_2=1, b_2=-3, c_2=2
\end{aligned}
$
Angle between the lines
$
\begin{array}{r}
\cos \theta=\left(\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right|\right) \\
\therefore \quad \theta=\cos ^{-1}\left(\left|\frac{2-3-6}{\sqrt{14} \sqrt{14}}\right|\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}
\end{array}
$