3D Geometry 3 Question 12
####12. The length of the perpendicular from the point $(2,-1,4)$ on the straight line, $\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}$ is
(a) greater than 3 but less than 4
(b) less than 2
(c) greater than 2 but less than 3
(d) greater than 4
Show Answer
Answer:
(a)
Solution:
- Equation of given line is
$ \frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}=r(\text { let }) $
Coordinates of a point on line (i) is
$ A(10 r-3,-7 r+2, r) $
Now, let the line joining the points $P(2,-1,4)$ and $A(10 r-3,-7 r+2, r)$ is perpendicular to line (i). Then,
$ \mathbf{P A} \cdot(10 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+\hat{\mathbf{k}})=0 $
$[\because$ vector along line (i) is $(10 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+\hat{\mathbf{k}})]$
$\Rightarrow[(10 r-5) \hat{\mathbf{i}}+(-7 r+3) \hat{\mathbf{j}}+(r-4) \hat{\mathbf{k}}] \cdot[10 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+\hat{\mathbf{k}}]=0$
$\Rightarrow 10(10 r-5)-7(3-7 r)+(r-4)=0$
$\Rightarrow 100 r-50-21+49 r+r-4=0$
$\Rightarrow 150 r=75 \Rightarrow r=\frac{1}{2}$
So, the foot of perpendicular is $A 2,-\frac{3}{2}, \frac{1}{2}$
[put $r=\frac{1}{2}$ in the coordinates of point $A$ ]
Now, perpendicular distance of point $P(2,-1,4)$ from the line (i) is
$ \begin{aligned} P A & =\sqrt{(2-2)^{2}+-\frac{3}{2}+1^{2}+\frac{1}{2}-4^{2}} \ & =\sqrt{\frac{1}{4}+\frac{49}{4}}=\sqrt{\frac{50}{4}}=\frac{5}{\sqrt{2}} \end{aligned} $
which lies in $(3,4)$.
