Let ( , , ) be the co-ordinates of the foot of the perpendicular drawn from the point (5,4…

Question

Let $(\alpha ,\beta ,\gamma )$ be the co-ordinates of the foot of the perpendicular drawn from the point $(5,4,2)$ on the line $\overset{	o }{r}=(-\hat{i}+3\hat{j}+\hat{k})+\lambda (2\hat{i}+3\hat{j}-\hat{k})$.

Then the length of the projection of the vector $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ on the vector $6\hat{i}+2\hat{j}+3\hat{k}$ is :

Accepted Answer

$\begin{gathered} \frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}=\lambda \ (2\lambda -1,3\lambda +3,1-\lambda ) \ \overset{	o \leftarrow }{PF}=(2\lambda -6)\hat{i}+(3\lambda -1)\hat{j}+(-\lambda -1)\hat{k} \ \overset{	o \leftarrow }{PF}\cdot (2\hat{i}+3\hat{j}-\hat{k})=0 \ 4\lambda -12+9\lambda -3+\lambda +1=0\Rightarrow \lambda =1 \ \Rightarrow F\equiv (1,6,0) \ \Rightarrow \hat{i}+6\hat{j}6\hat{i}+2\hat{j}+3\hat{k} \ =|\frac{(\hat{i}+6\hat{j})\cdot (6\hat{i}+2\hat{j}+3\hat{k})}{|6\hat{i}+2\hat{j}+3\hat{k}|}| \ =\frac{1}{7}(6+12)=\frac{18}{7} \ \end{gathered}$

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