Question

If $\bar{a} = \hat{i} + 2\hat{j} - 3\hat{k}$, $\bar{b} = 3\hat{i} - \hat{j} + 2\hat{k}$, $\bar{c} = \hat{i} + 3\hat{j} + \hat{k}$ and $\bar{a} + \lambda\bar{b}$ is perpendicular to $\bar{c}$, then $\lambda =$

• A. -2
• B. 4
• C. -4
• D. 2

Hint:

Whenever a problem asks for orthogonality involving a linear scalar addition like $\bar{a} + \lambda\bar{b} \perp \bar{c}$, immediately use the distributed shortcut: $\lambda = -\frac{\bar{a} \cdot \bar{c}}{\bar{b} \cdot \bar{c}}$.
This avoids having to write out the full component vector equation for $(\bar{a} + \lambda\bar{b})$ beforehand.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to find the scalar constant value $\lambda$ such that a linear combination of vectors $(\bar{a} + \lambda\bar{b})$ is geometrically orthogonal (perpendicular) to a third given vector $\bar{c}$.

Step 2: Key Formula or Approach:
Two non-zero vectors $\bar{u}$ and $\bar{v}$ are perpendicular if and only if their standard dot product evaluates to exactly zero:
$$\bar{u} \cdot \bar{v} = 0$$ Setting up our condition gives:
$$(\bar{a} + \lambda\bar{b}) \cdot \bar{c} = 0$$ By applying the distributive property of vector dot products, this can be split into:
$$(\bar{a} \cdot \bar{c}) + \lambda(\bar{b} \cdot \bar{c}) = 0 \implies \lambda = -\frac{\bar{a} \cdot \bar{c}}{\bar{b} \cdot \bar{c}}$$

Step 3: Detailed Explanation:
Let's compute the individual scalar dot products step-by-step using component multiplication ($u_x v_x + u_y v_y + u_z v_z$):
1. Compute $\bar{a} \cdot \bar{c}$:
$$\bar{a} \cdot \bar{c} = (1)(1) + (2)(3) + (-3)(1) = 1 + 6 - 3 = 4$$ 2. Compute $\bar{b} \cdot \bar{c}$:
$$\bar{b} \cdot \bar{c} = (3)(1) + (-1)(3) + (2)(1) = 3 - 3 + 2 = 2$$ Now, substitute these scalar product results back into the orthogonality condition:
$$4 + \lambda(2) = 0$$ $$2\lambda = -4 \implies \lambda = -2$$

Step 4: Final Answer:
The value of the scalar modifier $\lambda$ is -2, which corresponds to option (A).