Let A and B be two distinct points on the lin | vedclass

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

Question

(B) Let the line $L$ be $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} = \lambda$. Any point on the line is $Q(3\lambda+6, 2\lambda+7, -2\lambda+7)$.

Vedclass · Accepted Answer

$47$

Vedclass · Answer

(B) Let the line $L$ be $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} = \lambda$. Any point on the line is $Q(3\lambda+6, 2\lambda+7, -2\lambda+7)$.Let $P = (1, 2, 3)$. The vector $\overrightarrow{PQ} = (3\lambda+5, 2\lambda+5, -2\lambda+4)$.Since $PQ$ is perpendicular to the line $L$ with direction vector $\vec{b} = (3, 2, -2)$,we have $\overrightarrow{PQ} \cdot \vec{b} = 0$.$3(3\lambda+5) + 2(2\lambda+5) - 2(-2\lambda+4) = 0$$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0$$17\lambda + 17 = 0 \Rightarrow \lambda = -1$.Thus,the foot of the perpendicular is $Q(3(-1)+6, 2(-1)+7, -2(-1)+7) = Q(3, 5, 9)$.Points $A$ and $B$ are at a distance $d = 2\sqrt{17}$ from $Q$ on the line. Let the unit vector along the line be $\hat{u} = \frac{1}{\sqrt{3^2+2^2+(-2)^2}}(3, 2, -2) = \frac{1}{\sqrt{17}}(3, 2, -2)$.The points $A$ and $B$ are given by $Q \pm d\hat{u} = (3, 5, 9) \pm 2\sqrt{17} \cdot \frac{1}{\sqrt{17}}(3, 2, -2) = (3, 5, 9) \pm 2(3, 2, -2)$.$A = (3+6, 5+4, 9-4) = (9, 9, 5)$ and $B = (3-6, 5-4, 9+4) = (-3, 1, 13)$.Then $\overrightarrow{OA} \cdot \overrightarrow{OB} = (9, 9, 5) \cdot (-3, 1, 13) = (9)(-3) + (9)(1) + (5)(13) = -27 + 9 + 65 = 47$.

Similar Questions

Find the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$.

If a point $R(4, y, z)$ lies on the line joining the points $P(2, -3, 4)$ and $Q(8, 0, 10)$,then the distance of $R$ from the origin is

The Cartesian equation of the line which is parallel to the vector $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$ and passes through the point $(5, 2, -4)$ is . . . . . . .

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}$ and $\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{3}}$,then the sum of possible values of $\lambda$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module