If the direction ratios of the lines L_1 and | vedclass

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

Question

(B) Let the direction ratios of the line perpendicular to both $L_1$ and $L_2$ be $(a, b, c)$. Since the line is perpendicular to both,its direction r

Vedclass · Accepted Answer

$\pm \frac{1}{\sqrt{35}}, \pm \frac{5}{\sqrt{35}}, \pm \frac{3}{\sqrt{35}}$

Vedclass · Answer

(B) Let the direction ratios of the line perpendicular to both $L_1$ and $L_2$ be $(a, b, c)$. Since the line is perpendicular to both,its direction ratios are given by the cross product of the direction ratios of $L_1$ and $L_2$: $(a, b, c) = (2, -1, 1) 	imes (3, -3, 4) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 1 \ 3 & -3 & 4 \end{vmatrix} = \hat{i}(-4 + 3) - \hat{j}(8 - 3) + \hat{k}(-6 + 3) = -1\hat{i} - 5\hat{j} - 3\hat{k}$. Thus,the direction ratios are $(-1, -5, -3)$ or $(1, 5, 3)$. The magnitude is $\sqrt{1^2 + 5^2 + 3^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$. The direction cosines are $\pm \frac{1}{\sqrt{35}}, \pm \frac{5}{\sqrt{35}}, \pm \frac{3}{\sqrt{35}}$.

If the direction ratios of the lines $L_1$ and $L_2$ are $2, -1, 1$ and $3, -3, 4$ respectively,then the direction cosines of a line that is perpendicular to both $L_1$ and $L_2$ are

Similar Questions

If the position vectors of three points $A, B, C$ are $\hat{i} + \hat{j} + \hat{k}$,$2\hat{i} + 3\hat{j} - 4\hat{k}$,and $7\hat{i} + 4\hat{j} + 9\hat{k}$ respectively,then find the unit vector perpendicular to the plane of triangle $ABC$.

$A$ unit vector coplanar with $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ and perpendicular to $\vec{c} = \hat{i} + \hat{j} + \hat{k}$ is

If $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{b} \times \vec{c} = \vec{a}$,and $a, b, c$ are the moduli of the vectors $\vec{a}, \vec{b}, \vec{c}$ respectively,then:

Let $\vec{b}=\hat{i}+\hat{j}+\lambda \hat{k}, \lambda \in R$. If $\vec{a}$ is a vector such that $\vec{a} \times \vec{b}=13 \hat{i}-\hat{j}-4 \hat{k}$ and $\vec{a} \cdot \vec{b}+21=0$,then $(\vec{b}-\vec{a}) \cdot(\hat{k}-\hat{j})+(\vec{b}+\vec{a}) \cdot(\hat{i}-\hat{k})$ is equal to

For any three vectors $\vec{a}, \vec{b}, \vec{c}$,the value of $\vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times (\vec{c} + \vec{a}) + \vec{c} \times (\vec{a} + \vec{b})$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module