Question

Let $\vec{a}=\hat{i}+2\hat{j}-3\hat{k}$ and $\vec{b}=\lambda\hat{j}+3\hat{k}$. If the projection of $\vec{a}$ on $\vec{b}$ is equal to the projection of $\vec{b}$ on $\vec{a}$, then the values of $\lambda$ are

• A. $\pm\sqrt{7}$
• B. $\pm\sqrt{3}$
• C. $\pm 5$
• D. $\pm 3$
• E. $\pm\sqrt{5}$

Hint:

Logic Tip: If the projection of A onto B equals the projection of B onto A, their magnitudes must be equal ($|A| = |B|$) or they must be completely perpendicular to each other ($A \cdot B = 0$).

Answer 1

Answer: (E)

Concept:
The scalar projection of vector $\vec{x}$ onto vector $\vec{y}$ is given by the formula $\frac{\vec{x} \cdot \vec{y}}{|\vec{y}|}$. Setting the mutual projections equal to each other establishes a direct relationship between the magnitudes of the two vectors.

Step 1: Write the equality condition for the projections.
We are given: $$\text{Projection of } \vec{a} \text{ on } \vec{b} = \text{Projection of } \vec{b} \text{ on } \vec{a}$$ $$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$$

Step 2: Simplify the equation.
Since the dot product is commutative ($\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$), we can divide both sides by the dot product (assuming it is non-zero). This simplifies to: $$\frac{1}{|\vec{b}|} = \frac{1}{|\vec{a}|} \implies |\vec{a}| = |\vec{b}|$$ The condition implies the two vectors must have identical magnitudes.

Step 3: Calculate the magnitude of vector a.
Using $\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$: $$|\vec{a}| = \sqrt{1^2 + 2^2 + (-3)^2}$$ $$|\vec{a}| = \sqrt{1 + 4 + 9} = \sqrt{14}$$

Step 4: Set up the magnitude equation for vector b.
Using $\vec{b} = 0\hat{i} + \lambda\hat{j} + 3\hat{k}$: $$|\vec{b}| = \sqrt{0^2 + \lambda^2 + 3^2} = \sqrt{\lambda^2 + 9}$$ Equating the magnitudes: $$\sqrt{14} = \sqrt{\lambda^2 + 9}$$

Step 5: Solve for $\lambda$.
Square both sides of the equation: $$14 = \lambda^2 + 9$$ $$\lambda^2 = 14 - 9 = 5$$ Take the square root to find the values: $$\lambda = \pm\sqrt{5}$$ Hence the correct answer is (E) $\pm\sqrt{5$}.