Vector Algebra Part 3

Concept:The foot of the perpendicular from a point to a line is found by solving for the parameter where the direction vector of the line is perpendicular to the vector from the point to the foot.Explanation:Parameterize the line: $\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1} = \lambda$.General point on line: $(2\lambda -1, 3\lambda +3, 1-\lambda)$.Vector from given point $P(5,4,2)$ to foot $F$: $\overrightarrow{PF} = (2\lambda -6)\hat{i} + (3\lambda -1)\hat{j} + (-\lambda -1)\hat{k}$.Since $\overrightarrow{PF}$ is perpendicular to the direction vector of the line $2\hat{i}+3\hat{j}-\hat{k}$, their dot product is zero: $(2\lambda -6)(2) + (3\lambda -1)(3) + (-\lambda -1)(-1) = 0$.Simplify: $4\lambda -12 +9\lambda -3 +\lambda +1 = 0 \Rightarrow 14\lambda -14 =0 \Rightarrow \lambda =1$.Thus foot $F$ is $(1,6,0)$, so $\alpha=1, \beta=6, \gamma=0$.The vector $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k} = \hat{i}+6\hat{j}$.Projection of this vector onto $6\hat{i}+2\hat{j}+3\hat{k}$: $\frac{(\hat{i}+6\hat{j})\cdot(6\hat{i}+2\hat{j}+3\hat{k})}{|6\hat{i}+2\hat{j}+3\hat{k}|} = \frac{1(6)+6(2)+0(3)}{\sqrt{6^2+2^2+3^2}} = \frac{6+12}{\sqrt{36+4+9}} = \frac{18}{\sqrt{49}} = \frac{18}{7}$.Answer:$\frac{18}{7}$ (Option A).