Let (,,) be the coordinates of the foot of the perpendicular drawn from the point (5,4,2) …

Question

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

A general point on the line is $F \equiv (-1+2\lambda,\ 3+3\lambda,\ 1-\lambda)$. Then $\vec{PF} = F - (5,4,2) = (2\lambda-6)\hat i + (3\lambda-1)\hat j + (-\lambda-1)\hat k$.

Since $PF$ is perpendicular to the line's direction $(2,3,-1)$:
$$2(2\lambda-6) + 3(3\lambda-1) - 1(-\lambda-1) = 0 \implies 4\lambda-12+9\lambda-3+\lambda+1=0 \implies 14\lambda = 14 \implies \lambda=1$$

So $F \equiv (1,6,0)$, i.e. $\alpha\hat i+\beta\hat j+\gamma\hat k = \hat i+6\hat j$.

$$\text{Projection} = \frac{(\hat i+6\hat j)\cdot(6\hat i+2\hat j+3\hat k)}{|6\hat i+2\hat j+3\hat k|} = \frac{6+12}{\sqrt{36+4+9}} = \frac{18}{7}$$

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