Three Dimensional Geometry Question 34
Question 34 - 01 February - Shift 2
The point of intersection $C$ of the plane $8x + y + 2z = 0$ and the line joining the points $A(–3, –6, 1)$ and $B(2, 4, –3)$ divides the line segment $AB$ internally in the ratio $k : 1.$ If $a, b, c (|a|, |b|, |c|$ are coprime $)$ are the direction ratios of the perpendicular from the point $C$ on the line $\frac{1-x}1 = \frac{y + 4}2 = \frac{z + 2}3$, then $|a + b + c|$ is equal to ______.
Show Answer
Answer: 10
Solution:
Formula: Section Formula, Direction Cosines And Direction Ratios
Plane : $8 x+y+2 z=0$
Given line $A B: \frac{x-2}{5}=\frac{y-4}{10}=\frac{z+3}{-4}=\lambda$
Any point on line $(5 \lambda+2,10 \lambda+4,-4 \lambda-3)$
Point of intersection of line and plane
$8(5 \lambda+2)+10 \lambda+4-8 \lambda-6=0$
$\lambda=-\frac{1}{3}$
C $(\frac{1}{3}, \frac{2}{3},-\frac{5}{3})$
$L: \frac{x-1}{-1}=\frac{y+4}{2}=\frac{z+2}{3}=\mu$
$\overrightarrow{{}CD}=(-\mu+\frac{2}{3}) \hat{i}+(2 \mu-\frac{14}{3}) \hat{j}+(3 \mu-\frac{1}{3}) \hat{k}$
$(-\mu+\frac{2}{3})(-1)+(2 \mu-\frac{14}{3}) 2+(3 \mu-\frac{1}{3}) 3=0$
$\mu=\frac{11}{14}$
$\overrightarrow{{}CD}=\frac{-5}{42}, \frac{-130}{42}, \frac{85}{42}$
Direction ratios $(-1,-26,17)$
$\therefore |a+b+c| = |-1-26+17|$
$|a+b+c|=|-10|$
$|a+b+c|=10$
BETA
