Question

Let a plane P contain two lines
\(\vec{r} = \hat{i} + \lambda(\hat{i} + \hat{j})\), \(\lambda \in R\) and
\(\vec{r} = -\hat{j} + \mu(\hat{j} - \hat{k})\), \(\mu \in R\).
If \(Q(\alpha, \beta, \gamma)\) is the foot of the perpendicular drawn from the point M(1, 0, 1) to P, then \(3(\alpha + \beta + \gamma)\) equals ______ .

Hint:

The normal of P is the cross product of the two direction vectors; the foot of the perpendicular is M plus a multiple of the normal.

Answer 1

Correct Answer: 5

Line 1: passes through (1, 0, 0), direction \(\vec{d_1} = (1, 1, 0)\). Line 2: passes through (0, -1, 0), direction \(\vec{d_2} = (0, 1, -1)\).
Normal \(\vec{n} = \vec{d_1} \times \vec{d_2} = (1\cdot(-1) - 0\cdot1,\; 0\cdot0 - 1\cdot(-1),\; 1\cdot1 - 1\cdot0) = (-1, 1, 1)\).
Plane through (1, 0, 0): \(-(x-1) + y + z = 0\), i.e. \(-x + y + z + 1 = 0\). (Check (0, -1, 0): \(0 - 1 + 0 + 1 = 0\).)
For M(1, 0, 1): value \(= -1 + 0 + 1 + 1 = 1\), \(|\vec{n}|^2 = 3\).
Foot: \(Q = M - \frac{1}{3}\vec{n} = (1 + \frac13,\; -\frac13,\; 1 - \frac13) = (\frac43, -\frac13, \frac23)\).
\(\alpha + \beta + \gamma = \frac{4 - 1 + 2}{3} = \frac53\), so \(3(\alpha + \beta + \gamma) = 5\).
Answer: 5