If the lines frac{x-3}{2}=frac{y-2}{3}=frac{z | vedclass

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

Question

(C) The condition for two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ to

Vedclass · Accepted Answer

$3\pi -8$

Vedclass · Answer

(C) The condition for two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ to be coplanar is $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} = 0$.For the given lines,$(x_1, y_1, z_1) = (3, 2, 1)$ and $(x_2, y_2, z_2) = (2, 3, 2)$.The direction ratios are $(a_1, b_1, c_1) = (2, 3, \lambda)$ and $(a_2, b_2, c_2) = (3, 2, 3)$.The vector $\vec{AB} = (2-3, 3-2, 2-1) = (-1, 1, 1)$.Substituting these into the determinant:$\begin{vmatrix} -1 & 1 & 1 \ 2 & 3 & \lambda \ 3 & 2 & 3 \end{vmatrix} = 0$.Expanding the determinant: $-1(9-2\lambda) - 1(6-3\lambda) + 1(4-9) = 0$.$-9 + 2\lambda - 6 + 3\lambda - 5 = 0 \Rightarrow 5\lambda - 20 = 0 \Rightarrow \lambda = 4$.Now,we need to evaluate $\sin ^{-1}(\sin 4) + \cos ^{-1}(\cos 4)$.Since $4$ radians is in the third quadrant $(\pi $\sin ^{-1}(\sin 4) = \sin ^{-1}(\sin(\pi - 4)) = \pi - 4$.$\cos ^{-1}(\cos 4) = \cos ^{-1}(\cos(2\pi - 4)) = 2\pi - 4$.Adding these: $(\pi - 4) + (2\pi - 4) = 3\pi - 8$.

If the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{\lambda}$ and $\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-2}{3}$ are coplanar,then $\sin ^{-1}(\sin \lambda)+\cos ^{-1}(\cos \lambda)=$

Similar Questions

The equation of the line passing through the point of intersection of $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ and also through the point $(2,1,-2)$ is

The lines $\frac{6x-6}{18} = \frac{y+1}{3} = \frac{z-1}{5}$ and $\frac{3x+6}{12} = \frac{y-1}{3} = \frac{z+1}{2}$ are $\dots$

The graph of the equation $y^2 + z^2 = 0$ in three-dimensional space is

Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is

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