The equation of a line passing through the po | vedclass

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

Question

(B) 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

$\frac{x-2}{2}=\frac{1-y}{7}=\frac{z-3}{4}$

Vedclass · Answer

(B) 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 $\langle 1, 2, 3 \rangle$ and $\langle -3, 2, 5 \rangle$,we have: $1a + 2b + 3c = 0$ $-3a + 2b + 5c = 0$ Using the cross product to find the direction ratios $\langle a, b, c \rangle$: $a = (2)(5) - (3)(2) = 10 - 6 = 4$ $b = (3)(-3) - (1)(5) = -9 - 5 = -14$ $c = (1)(2) - (2)(-3) = 2 + 6 = 8$ Thus,the direction ratios are $\langle 4, -14, 8 \rangle$,which simplifies to $\langle 2, -7, 4 \rangle$. The equation of the line passing through $(2, 1, 3)$ with direction ratios $\langle 2, -7, 4 \rangle$ is: $\frac{x-2}{2} = \frac{y-1}{-7} = \frac{z-3}{4}$ This can be rewritten as: $\frac{x-2}{2} = \frac{1-y}{7} = \frac{z-3}{4}$

The equation of a line passing through the point $(2,1,3)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$ is

Similar Questions

The length and foot of the perpendicular from the point $(2, -1, 5)$ to the line $\frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11}$ are

Find the shortest distance between the lines whose vector equations are $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})$ and $\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$.

If $2x - y + z = 0 = y - x + 2z = mx - 2y + mz$ represents a line in space,then the value of $m$ is-

Vedclass Test Series

Exam Paper Generator

Online Exam Module