Question

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

• A. 5
• B. 2
• C. 3
• D. 4

Hint:

Bypassing full vector distribution saves precious time! Compute the scalar dot products $\vec{a}\cdot\vec{c}$ and $\vec{b}\cdot\vec{c}$ individually as simple integers right away. Then, plug them directly into the linear equation $\vec{a}\cdot\vec{c} + \lambda(\vec{b}\cdot\vec{c}) = 0$ to isolate $\lambda$ in single mental math step.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$. We need to find the scalar constant $\lambda$ such that the linear combination vector $\vec{a} + \lambda \vec{b}$ is perpendicular to vector $\vec{c}$.

Step 2: Key Formula or Approach:
Two non-zero vectors are perpendicular (orthogonal) if and only if their vector dot product is exactly equal to zero: $$(\vec{a} + \lambda \vec{b}) \cdot \vec{c} = 0$$ Using the distributive property of dot products, this can be expanded as: $$\vec{a} \cdot \vec{c} + \lambda (\vec{b} \cdot \vec{c}) = 0 \implies \lambda = -\frac{\vec{a} \cdot \vec{c}}{\vec{b} \cdot \vec{c}}$$

Step 3: Detailed Explanation:
1. Let's compute the dot product $\vec{a} \cdot \vec{c}$: $$\vec{a} \cdot \vec{c} = (1)(3) + (2)(1) + (3)(0) = 3 + 2 + 0 = 5$$ 2. Let's compute the dot product $\vec{b} \cdot \vec{c}$: $$\vec{b} \cdot \vec{c} = (-1)(3) + (2)(1) + (1)(0) = -3 + 2 + 0 = -1$$ 3. Substitute these dot product values into our orthogonality condition: $$5 + \lambda (-1) = 0$$ $$5 - \lambda = 0 \implies \lambda = 5$$

Step 4: Final Answer:
The value of $\lambda$ is 5, which corresponds to option (A).