If the distances of the point P(1, 2, a) from | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the line $L$ be $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} = k$. Any point on $L$ is $(k+1, 2k, k+1)$.Let $A$ be the intersection of $L_1$ and $

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Let the line $L$ be $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} = k$. Any point on $L$ is $(k+1, 2k, k+1)$.Let $A$ be the intersection of $L_1$ and $L$. For $L_1: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b} = \lambda$, a point on $L_1$ is $(3\lambda+1, 4\lambda+2, b\lambda+a)$.Since $A$ lies on $L$, we have $\frac{3\lambda+1-1}{1} = \frac{4\lambda+2}{2} = \frac{b\lambda+a-1}{1}$.From $\frac{3\lambda}{1} = \frac{4\lambda+2}{2}$, we get $6\lambda = 4\lambda+2 \Rightarrow 2\lambda = 2 \Rightarrow \lambda = 1$.Then $A = (3(1)+1, 4(1)+2, b(1)+a) = (4, 6, a+b)$.Since $A$ is on $L$, $\frac{4-1}{1} = \frac{6}{2} = \frac{a+b-1}{1} \Rightarrow 3 = 3 = a+b-1 \Rightarrow a+b=4$.Let $B$ be the intersection of $L_2$ and $L$. For $L_2: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c} = \mu$, a point on $L_2$ is $(\mu+1, 4\mu+2, c\mu+a)$.Since $B$ lies on $L$, $\frac{\mu+1-1}{1} = \frac{4\mu+2}{2} = \frac{c\mu+a-1}{1}$.From $\mu = \frac{4\mu+2}{2}$, we get $2\mu = 4\mu+2 \Rightarrow -2\mu = 2 \Rightarrow \mu = -1$.Then $B = (-1+1, 4(-1)+2, c(-1)+a) = (0, -2, a-c)$.Since $B$ is on $L$, $\frac{0-1}{1} = \frac{-2}{2} = \frac{a-c-1}{1} \Rightarrow -1 = -1 = a-c-1 \Rightarrow a-c=0 \Rightarrow a=c$.Given $PA = PB$, where $P(1, 2, a)$, $A(4, 6, a+b)$, and $B(0, -2, a-c)$.$PA^2 = (4-1)^2 + (6-2)^2 + (a+b-a)^2 = 3^2 + 4^2 + b^2 = 9+16+b^2 = 25+b^2$.$PB^2 = (0-1)^2 + (-2-2)^2 + (a-c-a)^2 = (-1)^2 + (-4)^2 + (-c)^2 = 1+16+c^2 = 17+c^2$.Since $PA=PB$, $25+b^2 = 17+c^2 \Rightarrow c^2 - b^2 = 8$.Since $a=c$ and $a+b=4$, we have $c+b=4$. Then $c-b = \frac{c^2-b^2}{c+b} = \frac{8}{4} = 2$.Solving $c+b=4$ and $c-b=2$, we get $2c=6 \Rightarrow c=3$ and $b=1$.Then $a=c=3$. Thus, $a+b+c = 3+1+3 = 7$.

If the distances of the point $P(1, 2, a)$ from the line $L: \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $L_{1}: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $L_{2}: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equal, then $a+b+c$ is equal to

Similar Questions

The shortest distance between the skew lines $r = (-\hat{i} + 3\hat{k}) + t(2\hat{i} + 3\hat{j} + 6\hat{k})$ and $r = (3\hat{i} + \hat{j} - \hat{k}) + s(2\hat{i} - \hat{j} + 2\hat{k})$ is

Find the angle between the following pair of lines:
$\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$

Lines $\frac{1-x}{3}=\frac{y-2}{1}=\frac{z-1}{2}$ and $\frac{x-2}{p}=\frac{y-1}{2}=\frac{z-2}{1}$ are mutually perpendicular to each other,then $p=$ . . . . . . .

The point having position vector $\vec{p} = 4\hat{i} - 11\hat{j} + 2\hat{k}$ lies on which of the following lines?

If lines $\frac{2x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}$ and $\frac{x-1}{1}=\frac{3y-1}{\lambda}=\frac{z-2}{1}$ are perpendicular to each other,then $\lambda = \ldots$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module