Question

The perpendicular distance of point $(2,-1,4)$ from the line $\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}$ lies between

• A. $(2,3)$
• B. $(3,4)$
• C. $(4,5)$
• D. $(1,2)$

Answer 1

The correct option is B

$(3,4)$

Explanation for the correct option:

Find the point of perpendicular distance from the given point

Let $P$ be $(2,-1,4)$.

$\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}=\lambda$

Direction cosine of line $10\hat{i}-7{\hat{j}}^{}+\hat{k}$

Let the foot of the perpendicular from $P$$(2,-1,4)$ to the given line be $A$$(10\lambda -3,-7\lambda +2,\lambda )$

$\overrightarrow{PA}·(10\hat{i}-7{\hat{j}}^{}+\hat{k})=0$
$\Rightarrow 10(10\lambda -5)-7(-7\lambda +3)+1(\lambda -4)=0$
$$\begin{gathered} \Rightarrow 150\lambda =75 \\ \Rightarrow \lambda =\frac{1}{2} \end{gathered}$$
$$\begin{gathered} \stackrel{\rightharpoonup }{\left|PA\right|}=\sqrt{{(10\lambda -5)}^2+{(-7\lambda +3)}^2+{(\lambda -4)}^2} \\ =\sqrt{0+{\left(\frac{1}{2}\right)}^2+{\left(\frac{7}{2}\right)}^2} \\ =\sqrt{\frac{50}{4}} \\ =3.53 \end{gathered}$$

$3.53$ lies in between $(3,4)$

Hence, the correct option is (B)