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

(D) Let the direction ratios of the required line be $\langle a, b, c \rangle$. Since the line is perpendicular to the lines with direction ratios $\l

Vedclass · Accepted Answer

$\langle \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \rangle$

Vedclass · Answer

(D) Let the direction ratios of the required line be $\langle a, b, c \rangle$. Since the line is perpendicular to the lines with direction ratios $\langle 1, -2, -2 \rangle$ and $\langle 0, 2, 1 \rangle$,we have: $1(a) - 2(b) - 2(c) = 0$  (Equation $1$) $0(a) + 2(b) + 1(c) = 0$  (Equation $2$) From Equation $2$,we get $c = -2b$. Substituting $c = -2b$ into Equation $1$: $a - 2b - 2(-2b) = 0$ $a - 2b + 4b = 0$ $a + 2b = 0 \implies a = -2b$. Let $b = -1$,then $a = 2$ and $c = 2$. The direction ratios are $\langle 2, -1, 2 \rangle$. The magnitude is $\sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$. The direction cosines are $\langle \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \rangle$.

The direction cosines of the line which is perpendicular to the lines with direction cosines proportional to $\langle 1, -2, -2 \rangle$ and $\langle 0, 2, 1 \rangle$ is given by

Similar Questions

If the direction cosines of two lines are such that $l+m+n=0$ and $l^2+m^2-n^2=0$,then the angle between them is

If the direction cosines of two lines are given by $l+m+n=0$ and $l^2-5m^2+n^2=0$,then the angle between them is

If $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are the direction cosines of two lines satisfying the relations $l^2+mn-6n^2=0$ and $2l-m+3n=0$,then $|l_1 l_2|+|m_1 m_2|=$

Let $A(1,-1,2), B(6,11,2), C(1,2,6)$ be three points. If $l_1, m_1, n_1$ are the direction cosines of $AB$ and $l_2, m_2, n_2$ are the direction cosines of $AC$,then $|l_1 l_2+m_1 m_2+n_1 n_2|=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module