The direction cosines of the line which is pe | vedclass

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

Question

(B) Let the direction ratios of the two lines be $\vec{v_1} = (2, -3, 1)$ and $\vec{v_2} = (1, 2, -2)$.Since the required line is perpendicular to bot

Vedclass · Accepted Answer

$\pm \frac{4}{\sqrt{90}}, \pm \frac{5}{\sqrt{90}}, \pm \frac{7}{\sqrt{90}}$

Vedclass · Answer

(B) Let the direction ratios of the two lines be $\vec{v_1} = (2, -3, 1)$ and $\vec{v_2} = (1, 2, -2)$.Since the required line is perpendicular to both lines,its direction ratios $(a, b, c)$ are given by the cross product $\vec{v_1} 	imes \vec{v_2}$.$\vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -3 & 1 \ 1 & 2 & -2 \end{vmatrix} = \hat{i}(6-2) - \hat{j}(-4-1) + \hat{k}(4+3) = 4\hat{i} + 5\hat{j} + 7\hat{k}$.Thus,the direction ratios are $(4, 5, 7)$.The magnitude is $\sqrt{4^2 + 5^2 + 7^2} = \sqrt{16 + 25 + 49} = \sqrt{90}$.The direction cosines are $\pm \frac{4}{\sqrt{90}}, \pm \frac{5}{\sqrt{90}}, \pm \frac{7}{\sqrt{90}}$.

The direction cosines of the line which is perpendicular to the lines $\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}$ and $\frac{x+5}{1}=\frac{y+3}{2}=\frac{z-6}{-2}$ are

Similar Questions

If $A(1,1,2)$, $B(4,2,1)$ and $C(2,3,5)$ are the vertices of a triangle, then a vector representing the median of the triangle through $A$ is

The angle between the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and $\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$ is

The Cartesian equation of the line which passes through the points $(3,1,2)$ and $(-1,2,1)$ is

The vector equation of the line $\frac{x+3}{2}=\frac{2y-3}{5}; z=-1$ is

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 2}{1} = \frac{y + m}{2} = \frac{z - 2}{1}$ intersect each other, then the value of $m$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module