Possible Value of Vector Dot Product with Sum Vector

To find a possible value of $(\vec{a} + \vec{b}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$, let us define the vector: $\vec{c} = 2\hat{i} + 3\hat{j} + 4\hat{k}$

We are given the vector cross-product relation: $\vec{a} \times \vec{c} = \vec{c} \times \vec{b}$

Using the anti-commutative property of the cross product, $\vec{c} \times \vec{b} = -\vec{b} \times \vec{c}$, we can rewrite the equation as: $\vec{a} \times \vec{c} = -\vec{b} \times \vec{c}$ $\implies \vec{a} \times \vec{c} + \vec{b} \times \vec{c} = \vec{0}$ $\implies (\vec{a} + \vec{b}) \times \vec{c} = \vec{0}$

This implies that the vector $(\vec{a} + \vec{b})$ is parallel to the vector $\vec{c}$. Hence, there exists a scalar $\lambda$ such that: $\vec{a} + \vec{b} = \lambda \vec{c} = \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$

Taking the magnitude on both sides: $|\vec{a} + \vec{b}| = |\lambda| |\vec{c}|$

We know that: $|\vec{c}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$

Given that $|\vec{a} + \vec{b}| = \sqrt{29}$, we substitute this into the magnitude equation: $\sqrt{29} = |\lambda| \sqrt{29} \implies |\lambda| = 1 \implies \lambda = \pm 1$

Therefore: $\vec{a} + \vec{b} = \pm (2\hat{i} + 3\hat{j} + 4\hat{k})$

Now, let $\vec{d} = -7\hat{i} + 2\hat{j} + 3\hat{k}$. We compute the dot product: $(\vec{a} + \vec{b}) \cdot \vec{d} = \pm (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$ $= \pm [ 2(-7) + 3(2) + 4(3) ]$ $= \pm [ -14 + 6 + 12 ]$ $= \pm 4$

Thus, the possible values of $(\vec{a} + \vec{b}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$ are $4$ or $-4$.

Among the given choices, the value $4$ is present in Option C.