Question

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

• A. $-14$
• B. $14$
• C. $2$
• D. $-2$

Hint:

Distributing the dot product before substituting the full vector components ($\vec{a}\cdot\vec{c} + \lambda(\vec{b}\cdot\vec{c}) = 0$) saves you from doing messy algebraic expansions with $\hat{i}, \hat{j}, \hat{k}$ components inside parentheses!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given three 3D vectors and a condition that a specific linear combination of the first two vectors is orthogonal (perpendicular) to the third vector. We must find the scalar multiplier $\lambda$.

Step 2: Key Formula or Approach:
Two non-zero vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if their dot product evaluates to exactly zero: $\vec{u} \cdot \vec{v} = 0$.
Therefore, we must solve the equation: $(\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0$.
Using the distributive property of the dot product, this expands to: $\vec{a} \cdot \vec{c} + \lambda(\vec{b} \cdot \vec{c}) = 0$.

Step 3: Detailed Explanation:
First, calculate the dot product $\vec{a} \cdot \vec{c}$:
$$\vec{a} \cdot \vec{c} = (2)(3) + (2)(1) + (3)(2)$$ $$\vec{a} \cdot \vec{c} = 6 + 2 + 6 = 14$$ Next, calculate the dot product $\vec{b} \cdot \vec{c}$:
$$\vec{b} \cdot \vec{c} = (-1)(3) + (2)(1) + (1)(2)$$ $$\vec{b} \cdot \vec{c} = -3 + 2 + 2 = 1$$ Now, substitute these scalar values back into our distributed equation:
$$14 + \lambda(1) = 0$$ $$\lambda = -14$$

Step 4: Final Answer:
The value of $\lambda$ is $-14$, corresponding to option (A).