Question

Let $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{b} = 3\hat{i} + \hat{j} + 6\hat{k}$, $\vec{c} = \hat{i} - \hat{j} - \hat{k}$ be three vectors. If a vector $\vec{x}$ satisfies $\vec{x} \times \vec{b} = \vec{b} \times \vec{c}$ and $\vec{x} . \vec{a} = 0$, then

$\vec{x} . (\hat{i} - \hat{j} + \hat{k}) =$

• A. 15
• B. 10
• C. 20
• D. 25

Answer 1

Answer: (A)

15

Answer 2

The condition $\vec{x} \times \vec{b} = \vec{b} \times \vec{c}$ can be rewritten as $(\vec{x} + \vec{c}) \times \vec{b} = \vec{0}$. This implies that $\vec{x} + \vec{c}$ is parallel to $\vec{b}$, so $\vec{x} + \vec{c} = \lambda \vec{b}$ for some scalar $\lambda$. Thus, $\vec{x} = \lambda \vec{b} - \vec{c}$.

Substituting this into the second condition, $\vec{x} . \vec{a} = 0$, we get $(\lambda \vec{b} - \vec{c}) . \vec{a} = 0$, which expands to $\lambda (\vec{b} . \vec{a}) - (\vec{c} . \vec{a}) = 0$.

Calculating the dot products: $\vec{b} . \vec{a} = (3)(2) + (1)(1) + (6)(-1) = 6 + 1 - 6 = 1$ $\vec{c} . \vec{a} = (1)(2) + (-1)(1) + (-1)(-1) = 2 - 1 + 1 = 2$

Substituting these values into the equation for $\lambda$: $\lambda (1) - 2 = 0 \implies \lambda = 2$.

Now we find $\vec{x}$: $\vec{x} = 2\vec{b} - \vec{c} = 2(3\hat{i} + \hat{j} + 6\hat{k}) - (\hat{i} - \hat{j} - \hat{k})$ $\vec{x} = (6\hat{i} + 2\hat{j} + 12\hat{k}) - (\hat{i} - \hat{j} - \hat{k})$ $\vec{x} = 5\hat{i} + 3\hat{j} + 13\hat{k}$

Finally, we calculate $\vec{x} . (\hat{i} - \hat{j} + \hat{k})$: $\vec{x} . (\hat{i} - \hat{j} + \hat{k}) = (5\hat{i} + 3\hat{j} + 13\hat{k}) . (\hat{i} - \hat{j} + \hat{k})$ $= (5)(1) + (3)(-1) + (13)(1) = 5 - 3 + 13 = 15$.