Question

Let \(\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}\) and \(\vec{b} = 2\hat{i} - 3\hat{j} + 5\hat{k}\). If \(\vec{r}\times\vec{a} = \vec{b}\times\vec{r}\),
\(\vec{r}\cdot\left(\alpha\hat{i} + 2\hat{j} + \hat{k}\right) = 3\) and \(\vec{r}\cdot\left(2\hat{i} + 5\hat{j} - \alpha\hat{k}\right) = -1\), \(\alpha\in\mathbb{R}\), then the value of \(\alpha + |\vec{r}|^2\) is equal to :

• A. 13
• B. 11
• C. 9
• D. 15

Hint:

\(\vec r\times\vec a=\vec b\times\vec r\) gives \(\vec r\times(\vec a+\vec b)=\vec 0\), so \(\vec r\) is parallel to \(\vec a+\vec b\).

Answer 1

The Correct Option is D

\(\vec{r}\times\vec{a} = \vec{b}\times\vec{r} = -\vec{r}\times\vec{b}\Rightarrow \vec{r}\times(\vec{a}+\vec{b}) = \vec{0}\).
\(\vec{a}+\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}\), so \(\vec{r} = \lambda(3\hat{i} - \hat{j} + 2\hat{k})\).
\(\vec{r}\cdot(\alpha\hat{i}+2\hat{j}+\hat{k}) = \lambda(3\alpha - 2 + 2) = 3\alpha\lambda = 3\Rightarrow \alpha\lambda = 1\).
\(\vec{r}\cdot(2\hat{i}+5\hat{j}-\alpha\hat{k}) = \lambda(6 - 5 - 2\alpha) = \lambda - 2\alpha\lambda = -1\).
Using \(\alpha\lambda = 1\): \(\lambda - 2 = -1\Rightarrow\lambda = 1\), hence \(\alpha = 1\).
\(\vec{r} = 3\hat{i} - \hat{j} + 2\hat{k}\), \(|\vec{r}|^2 = 9 + 1 + 4 = 14\).
\(\alpha + |\vec{r}|^2 = 1 + 14 = 15\).
Answer: (4) 15