Question

Which of the following is "CORRECT" option?
$$\begin{array}{cc}\text{List I}& \text{List II}\\ \text{(a) The coordinates of a point on the line}& \\ x=4y+5,z=3y-6\text{at a distance}3\text{from}& \\ \text{the point}(5,3,-6)\text{is/are}& \text{(p)}(-1,-2,0)\\ \text{(b) The plane containing the lines}& \\ \frac{x-2}{3}=\frac{y+3}{5}=\frac{z+5}{7}& \\ \text{and parallel to}\hat{i}+4\hat{j}+7\hat{k}\text{has}& \\ & \text{(q)}(5,0,-6)\\ \text{(c) A line passes through two points}A(2,-3-,1)& \\ \text{and}B(8,-1,2)\text{. The coordinates of a point}& \\ \text{on this line farthest to the origin and at a}& \\ \text{distance of}14\text{units from}A\text{is}& \text{(r)}(2,5,7)\\ \text{(d) The coordinates of the foot of the perpendicular}& \\ \text{from the point}(3,-1,11)\text{on the line}& \\ \frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}\text{is/are}& \text{(s)}(14,1,5)\end{array}$$

• A. $\text{a}\to \text{q}$
• B. $\text{c}\to \text{p}$
• C. $\text{b}\to \text{s}$
• D. $\text{d}\to \text{q}$

Answer 1

The correct option is A $\text{a}\to \text{q}$

$\text{(}a)$
Any point on the given line,
$(4\lambda +5,\lambda ,3\lambda -6)$
The distance between
$(4\lambda +5,\lambda ,3\lambda -6)$ and $(5,3,-6)$ is $3$, so
$(4\lambda +5-5{)}^2+(\lambda -3{)}^2+(3\lambda +6-6{)}^2=9$
$\Rightarrow 16{\lambda }^2+{\lambda }^2+9-6\lambda +9{\lambda }^2=9$
$\Rightarrow 26{\lambda }^2-6\lambda =0$
$\Rightarrow \lambda =0,\frac{3}{13}$
The point will be $(5,0,-6)$
$a\to q$

$\text{(b)}$
The equation of the plane containing the lines $\frac{x-2}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and parallel to $\hat{i}+4\hat{j}+7\hat{k}$ is $\left|\begin{array}{ccc}x-2& y+3& z+5\\ 1& 4& 7\\ 3& 5& 7\end{array}\right|=0$
$\Rightarrow x-2y+z-3=0$
Point $(-1,-2,0)$ lies on this plane.
$b\to p$

$\text{(c)}$
A line passing through points $A(2,-3,-1)$ and $B(8,-1,2)$ is,
$\frac{x-2}{8-2}=\frac{y+3}{-1+3}=\frac{z+1}{2+1}$
$\Rightarrow \frac{x-2}{6}=\frac{y+3}{2}=\frac{z+1}{3}$
Any point on this line is of the form
$P(6\lambda +2,2\lambda -3,3\lambda -1)$
Distannce from point $A(2,-3,-1)$ is $14$ units.
Therefore,
$PA=14$
$P{A}^2=(14{)}^2$
$(6\lambda {)}^2+(2\lambda {)}^2+(3\lambda {)}^2=196$
$49{\lambda }^2=196$
${\lambda }^2=4$
$\lambda =\pm 2$
So the required points are $(14,1,5)$ and $(-10,-7,-7)$
The point farthest to the origin is $(14,1,5)$
$c\to s$

$\text{(d)}$
Any point on the line,
$\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ is,
$M(2\lambda ,3\lambda +2,4\lambda +3)$
$P(3,-1,11)$
Therefore, the directional ratios of $PM$,
$2\lambda -3,3\lambda +3,4\lambda -8$.
We know $PM$ is perpendicular to $AB$
$\Rightarrow 2(2\lambda -3)+3(3\lambda +3)+4(4\lambda -8)=0$
$4\lambda -6+9\lambda +9+16\lambda -32=0$
$\lambda =1$
Therefore, foot of the perpendicular is $M(2,5,7)$
$d\to r$