Question

If $\vec{a}\times(\hat{i}-\hat{j}+\hat{k})=(\hat{i}-\hat{j}+\hat{k})\times\vec{b}$ and $|\vec{a}+\vec{b}|=3\sqrt{3}$, then the possible values of $(\vec{a}+\vec{b})\cdot(3\hat{i}+2\hat{j}+\hat{k})$ are

• A. $\pm 1$
• B. $\pm 2$
• C. $\pm 4$
• D. $\pm 6$
• E. $\pm 8$

Hint:

Logic Tip: Rearranging cross products to equal $\vec{0}$ is a fundamental trick in vector algebra to prove collinearity/parallelism. Whenever you see $\vec{A} \times \vec{C} = \vec{C} \times \vec{B}$, immediately deduce that $(\vec{A}+\vec{B})$ is parallel to $\vec{C}$.

Answer 1

The Correct Option is D

Concept:
The cross product has an anti-commutative property: $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$. If the cross product of a vector with $\vec{v}$ is zero ($\vec{x} \times \vec{v} = 0$), it means the vector $\vec{x}$ is parallel to $\vec{v}$, so it can be written as a scalar multiple: $\vec{x} = \lambda\vec{v}$.
Step 1: Simplify the cross product equation.
Let $\vec{v} = \hat{i} - \hat{j} + \hat{k}$. The given equation is: $$\vec{a} \times \vec{v} = \vec{v} \times \vec{b}$$ Apply the anti-commutative property to the right side: $$\vec{a} \times \vec{v} = -(\vec{b} \times \vec{v})$$ Bring all terms to the left side: $$\vec{a} \times \vec{v} + \vec{b} \times \vec{v} = \vec{0}$$ Use the distributive property of the cross product: $$(\vec{a} + \vec{b}) \times \vec{v} = \vec{0}$$
Step 2: Express $(\vec{a}+\vec{b})$ as a scalar multiple of $\vec{v}$.
Since their cross product is the zero vector, the vector $(\vec{a} + \vec{b})$ must be parallel to $\vec{v}$. Let $\vec{a} + \vec{b} = \lambda\vec{v} = \lambda(\hat{i} - \hat{j} + \hat{k})$, where $\lambda$ is a scalar constant.
Step 3: Use the magnitude condition to solve for $\lambda$.
We are given $|\vec{a} + \vec{b}| = 3\sqrt{3}$. Substitute the expression from Step 2: $$|\lambda(\hat{i} - \hat{j} + \hat{k})| = 3\sqrt{3}$$ $$|\lambda| \cdot \sqrt{1^2 + (-1)^2 + 1^2} = 3\sqrt{3}$$ $$|\lambda| \cdot \sqrt{3} = 3\sqrt{3}$$ Divide by $\sqrt{3}$: $$|\lambda| = 3 \implies \lambda = \pm 3$$ Thus, $\vec{a} + \vec{b} = \pm 3(\hat{i} - \hat{j} + \hat{k})$.
Step 4: Compute the final dot product.
We need the value of $(\vec{a} + \vec{b}) \cdot (3\hat{i} + 2\hat{j} + \hat{k})$. Substitute our resolved vector: $$= [\pm 3(\hat{i} - \hat{j} + \hat{k})] \cdot (3\hat{i} + 2\hat{j} + \hat{k})$$ Calculate the dot product by multiplying corresponding components: $$= \pm 3 \left( (1)(3) + (-1)(2) + (1)(1) \right)$$ $$= \pm 3 (3 - 2 + 1)$$ $$= \pm 3 (2)$$ $$= \pm 6$$