Value of $\lambda$ such that the line $\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{\mathcal{z} - 1}{\lambda}$ is $\bot$ to normal to the plane $\overrightarrow{r} \bullet \left( 2\overrightarrow{i} + 3\overrightarrow{j} + 4\overrightarrow{k} \right) = 0$ is

Question

Dr. Abubacker · Accepted Answer

(a)
Since line is parallel to the plane vector, $2\overrightarrow{i} + 3\overrightarrow{j} + \lambda\overrightarrow{k}$ is perpendicular to the normal to the plane
$2\overrightarrow{i} + 3\overrightarrow{j} + 4\overrightarrow{k}$
$\Rightarrow 2 	imes 2 + 3 	imes 3 + 4\lambda = 0$
$\Rightarrow \ \lambda = - \frac{13}{4}$

Q. Value of $\lambda$ such that the line $\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{\mathcal{z} - 1}{\lambda}$ is $\bot$ to normal to the plane $\overrightarrow{r} \bullet \left( 2\overrightarrow{i} + 3\overrightarrow{j} + 4\overrightarrow{k} \right) = 0$ is

Explanation

Reviewed by Dr. Abubacker

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