Let P be the point (10, -2, -1) and Q be the | vedclass

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

Question

(A) The line passes through $A(2, -5, 11)$ and $B(-6, 7, -5)$.The direction ratios of the line are $(-6-2, 7-(-5), -5-11) = (-8, 12, -16)$.Dividing by

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(A) The line passes through $A(2, -5, 11)$ and $B(-6, 7, -5)$.The direction ratios of the line are $(-6-2, 7-(-5), -5-11) = (-8, 12, -16)$.Dividing by $-4$,the simplified direction ratios are $(2, -3, 4)$.The equation of the line is $\frac{x-2}{2} = \frac{y+5}{-3} = \frac{z-11}{4} = \lambda$.Any point $Q$ on the line is $(2\lambda+2, -3\lambda-5, 4\lambda+11)$.The vector $\vec{RQ} = (2\lambda+2-1, -3\lambda-5-7, 4\lambda+11-6) = (2\lambda+1, -3\lambda-12, 4\lambda+5)$.Since $RQ$ is perpendicular to the line,the dot product of $\vec{RQ}$ and the direction vector $(2, -3, 4)$ is $0$:$2(2\lambda+1) - 3(-3\lambda-12) + 4(4\lambda+5) = 0$$4\lambda + 2 + 9\lambda + 36 + 16\lambda + 20 = 0$$29\lambda + 58 = 0 \Rightarrow \lambda = -2$.Substituting $\lambda = -2$ into the coordinates of $Q$:$Q = (2(-2)+2, -3(-2)-5, 4(-2)+11) = (-2, 1, 3)$.The length $PQ$ where $P(10, -2, -1)$ and $Q(-2, 1, 3)$ is:$PQ = \sqrt{(-2-10)^2 + (1-(-2))^2 + (3-(-1))^2}$$PQ = \sqrt{(-12)^2 + (3)^2 + (4)^2} = \sqrt{144 + 9 + 16} = \sqrt{169} = 13$.

Let $P$ be the point $(10, -2, -1)$ and $Q$ be the foot of the perpendicular drawn from the point $R(1, 7, 6)$ on the line passing through the points $(2, -5, 11)$ and $(-6, 7, -5)$. Then the length of the line segment $PQ$ is equal to ..........

Similar Questions

Find the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$.

The shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ 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

If the lines $\frac{3-x}{2}=\frac{5y-2}{3\lambda+1}=5-z$ and $\frac{x+2}{-1}=\frac{1-3y}{7}=\frac{4-z}{2\mu}$ are at right angles,then $7\lambda-10\mu=$

If $A(1, 0, 2)$,$B(2, 1, 0)$,$C(2, -5, 3)$,and $D(0, 3, 2)$ are four points and the point of intersection of the lines $AB$ and $CD$ is $P(a, b, c)$,then $a + b + c =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module