Question

Let \(\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}\), \(\vec{b} = \hat{i} - \hat{j} + 2\hat{k}\), \(\vec{c} = 2\hat{i} + \hat{j} - 4\hat{k}\) be three vectors. If \(\vec{r}\) is the vector such that \(\vec{r} \times \vec{a} = (\vec{b} + \vec{c}) \times \vec{a}\) and \(\vec{r} \cdot (\vec{b} - \vec{c}) = 0\), then \(\vec{r} \cdot (\hat{i} + \hat{j} - \hat{k})\) is equal to :

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

\((\vec{r} - \vec{b} - \vec{c}) \times \vec{a} = \vec{0}\) means \(\vec{r} = \vec{b} + \vec{c} + \lambda\vec{a}\); then use the dot-product condition.

Answer 1

The Correct Option is C

From \(\vec{r} \times \vec{a} = (\vec{b} + \vec{c}) \times \vec{a}\) we get \((\vec{r} - \vec{b} - \vec{c}) \times \vec{a} = \vec{0}\), so \(\vec{r} = \vec{b} + \vec{c} + \lambda\vec{a}\).
\(\vec{b} + \vec{c} = (3, 0, -2)\), so \(\vec{r} = (3 + \lambda,\; 2\lambda,\; -2 + 3\lambda)\).
\(\vec{b} - \vec{c} = (-1, -2, 6)\). The condition \(\vec{r} \cdot (\vec{b} - \vec{c}) = 0\) gives \(-(3 + \lambda) - 4\lambda + 6(-2 + 3\lambda) = 0\), i.e. \(13\lambda - 15 = 0\), \(\lambda = \dfrac{15}{13}\).
Now \(\vec{r} \cdot (1, 1, -1) = (3 + \lambda) + 2\lambda - (-2 + 3\lambda) = 5\) (the \(\lambda\) terms cancel).
Answer: (3) 5