Let a line passing through the point P(4,1,0) | vedclass

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

Question

(A) Let the line $L_1$ be $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=p$. Then $A = (2p+1, 3p+2, 4p+3)$.Let the line $L_2$ be $\frac{x-6}{1}=\frac{y}{1

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Let the line $L_1$ be $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=p$. Then $A = (2p+1, 3p+2, 4p+3)$.Let the line $L_2$ be $\frac{x-6}{1}=\frac{y}{1}=\frac{z-4}{-1}=q$. Then $B = (q+6, q, 4-q)$.The points $P(4,1,0)$,$A$,and $B$ are collinear. The direction ratios of $PA$ are $(2p+1-4, 3p+2-1, 4p+3-0) = (2p-3, 3p+1, 4p+3)$.The direction ratios of $AB$ are $(q+6-(2p+1), q-(3p+2), 4-q-(4p+3)) = (q-2p+5, q-3p-2, -q-4p+1)$.Since $P, A, B$ are collinear,the direction ratios of $PA$ and $AB$ are proportional:$\frac{2p-3}{q-2p+5} = \frac{3p+1}{q-3p-2} = \frac{4p+3}{-q-4p+1} = k$.Solving this system,we find $p=-1$ and $q=3$.Substituting $p=-1$ into $A$,we get $A(-1, -1, -1)$.Substituting $q=3$ into $B$,we get $B(9, 3, 1)$.Now,calculate the determinant:$\left|\begin{array}{ccc}1 & 0 & 1 \ -1 & -1 & -1 \ 9 & 3 & 1\end{array}ight| = 1(-1 - (-3)) - 0 + 1(-3 - (-9)) = 1(2) + 1(6) = 8$.

Similar Questions

The shortest distance between the lines $\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}$ and $\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}$ is $......$.

The straight line $\frac{x - 3}{3} = \frac{y - 2}{1} = \frac{z - 1}{0}$ is

The vector equation of the line $2x+4=3y+1=6z-3$ is

The angle between the lines $2x = 3y = -z$ and $6x = -y = -4z$ is (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module