Question

If the vectors, \(\vec{p} = (a+1)\hat{i} + a\hat{j} + a\hat{k}\), \(\vec{q} = a\hat{i} + (a+1)\hat{j} + a\hat{k}\) and \(\vec{r} = a\hat{i} + a\hat{j} + (a+1)\hat{k}\) \((a \in R)\) are coplanar and \(3(\vec{p}\cdot\vec{q})^2 - \lambda|\vec{r}\times\vec{q}|^2 = 0\), then the value of \(\lambda\) is ______.

Hint:

Coplanarity means the scalar triple product is zero, which fixes \(a\); then evaluate \((\vec p\cdot\vec q)^2\) and \(|\vec r\times\vec q|^2\).

Answer 1

Correct Answer: 1

Coplanar: \(\begin{vmatrix}a+1&a&a\\a&a+1&a\\a&a&a+1\end{vmatrix} = 1 + 3a = 0 \Rightarrow a = -\frac13\).
Then \(\vec p = \frac13(2,-1,-1)\), \(\vec q = \frac13(-1,2,-1)\), \(\vec r = \frac13(-1,-1,2)\).
\(\vec p\cdot\vec q = \frac19(-2-2+1) = -\frac13\), so \(3(\vec p\cdot\vec q)^2 = 3\cdot\frac19 = \frac13\).
\(|\vec r|^2 = |\vec q|^2 = \frac{6}{9} = \frac23\), \(\vec r\cdot\vec q = \frac19(1-2-2) = -\frac13\).
\(|\vec r\times\vec q|^2 = |\vec r|^2|\vec q|^2 - (\vec r\cdot\vec q)^2 = \frac49 - \frac19 = \frac13\).
The condition gives \(\frac13 - \lambda\cdot\frac13 = 0 \Rightarrow \lambda = 1\).
Answer: 1