Question

The foot of perpendicular of the point $(2,0,5)$ on the line $\frac{x+1}{2}=\frac{y-1}{5}-\frac{z+1}{-1}$ is $(a, \beta, \gamma)$. Then which of the following is NOT correct?

• A. $\frac{\gamma}{\alpha}=\frac{5}{8}$
• B. $\frac{\beta}{\gamma}=-5$
• C. $\frac{\alpha \beta}{\gamma}=\frac{4}{15}$
• D. $\frac{\alpha}{\beta}=-8$

Answer 1

The Correct Option is B

To find the foot of the perpendicular, apply the formula for the point-line distance and the projection of a point onto a line using vector operations. Calculate \(\mathbf{PA} \cdot \mathbf{b} = 0\) where \(\mathbf{P}\) is the point on the line closest to \(\mathbf{A}\) and \(\mathbf{b}\) is the direction vector of the line. This provides the specific coordinates \(\alpha, \beta, \gamma\) of the point \(\mathbf{P}\), and subsequently, the ratios of these coordinates are compared against the options to identify the incorrect statement.

Answer 2

The correct answer is (B) : $\frac{\beta}{\gamma}=-5$
$L:\frac{x+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}=\lambda$

Let foot of perpendicular is
$P(2\lambda -1,5\lambda +1,-\lambda -1)$
$\overrightarrow{PA}=(3-2\lambda )\hat{i}-(5\lambda +1)\hat{j}+(6+\lambda )\hat{k}$
Direction ratio of line $\Rightarrow \vec{b}=2\hat{i}+5\hat{j}-\hat{k}$
$\text{Now,}\Rightarrow \overrightarrow{PA}\cdot \overrightarrow{b}=0$
$\Rightarrow 2(3-2\lambda )-5(5\lambda +1)-(6+\lambda )=0$
$\Rightarrow \lambda =\frac{-1}{6}$
$P(2\lambda -1,5\lambda +1,-\lambda -1)\equiv P(\alpha ,\beta ,\gamma )$
$\Rightarrow \alpha =2\left(-\frac{1}{6}\right)-1=-\frac{4}{3}\Rightarrow \alpha =-\frac{4}{3}$
$\Rightarrow \beta =5\left(-\frac{1}{6}\right)+1=\frac{1}{6}\Rightarrow \beta =\frac{1}{6}$
$\Rightarrow \gamma =-\lambda -1=\frac{1}{6}-1\Rightarrow \gamma =-\frac{5}{6}$