The coordinates of the foot of the perpendicu | vedclass

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

Question

(D) Let the given line be $\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z + 10}{8} = k$. Any point $L$ on the line is given by $L = (2k + 1, -3k - 1, 8k

Vedclass · Accepted Answer

$(3, -4, -2)$

Vedclass · Answer

(D) Let the given line be $\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z + 10}{8} = k$. Any point $L$ on the line is given by $L = (2k + 1, -3k - 1, 8k - 10)$. Let $P = (1, 0, 0)$. The direction ratios of the line $PL$ are $(2k + 1 - 1, -3k - 1 - 0, 8k - 10 - 0) = (2k, -3k - 1, 8k - 10)$. Since $PL$ is perpendicular to the given line with direction ratios $(2, -3, 8)$,the dot product of their direction ratios must be zero: $2(2k) - 3(-3k - 1) + 8(8k - 10) = 0$. $4k + 9k + 3 + 64k - 80 = 0$. $77k - 77 = 0$,which gives $k = 1$. Substituting $k = 1$ into the coordinates of $L$: $L = (2(1) + 1, -3(1) - 1, 8(1) - 10) = (3, -4, -2)$.

The coordinates of the foot of the perpendicular from the point $(1, 0, 0)$ to the line $\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z + 10}{8}$ are

Similar Questions

If the lines $\frac{x - 1}{k} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x - 2}{3} = \frac{y - 3}{k} = \frac{z - 1}{2}$ intersect,find the value of $k$.

The distance of the point $(1, 2, -4)$ from the line $\frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6}$ is

The Cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} .$ Write its vector form.

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

The shortest distance (in units) between the lines $\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\vec{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda(\hat{i}+2\hat{j})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module