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

The key idea is that two vectors are perpendicular when their dot product equals zero.

Step 1: Write the vector $\vec{a} + \lambda \vec{b}$.

Step 2: Apply the perpendicular condition: $(\vec{a} + \lambda \vec{b}) \cdot \vec{c} = 0$.

$[(2-\lambda)\hat{i} + (2+2\lambda)\hat{j} + (3+\lambda)\hat{k}] \cdot (3\hat{i} + \hat{j} + 0\hat{k}) = 0$ …

Two vectors are perpendicular exactly when their dot product is zero. That’s the entire engine of this problem. We’re told that $\vec{a}+\lambda\vec{b}$ is perpendicular to $\vec{c}$, so:

This single equation will let us solve for $\lambda$ because the dot product is linear — we can expand it into $\vec{a}\cdot\vec{c} + \lambda(\vec{b}\cdot\vec{c}) = 0$, then compute each dot product from the given components.

Let’s do it step by step.

1. Write the vectors clearly

   $\vec{a}=2\hat{i}+2\hat{j}+3\hat{k}$

   $\vec{b}=-\hat{i}+2\hat{j}+\hat{k}$

   $\vec{c}=3\hat{i}+\hat{j}+0\hat{k}$ (note the zero $\hat{k}$ component — easy to miss)

2. Set up the perpendicular condition

Expand using distributivity of the dot product:

3. Compute $\vec{a}\cdot\vec{c}$ Multiply corresponding components and add:

4. Compute $\vec{b}\cdot\vec{c}$

5. Plug into the equation

So $\lambda = 8$. …

Method: Solving for a Scalar via the Perpendicularity Condition

Use this when a combination of vectors carrying an unknown scalar must be perpendicular to another vector.

Steps

Step 1: Turn 'perpendicular' into a dot product equal to zero.

Step 2: Expand using linearity of the dot product.

For $\vec{p}=\vec{a}+\lambda\vec{b}$ and $\vec{q}=\vec{c}$,

$(\vec{a}+\lambda\vec{b})\cdot\vec{c} = \vec{a}\cdot\vec{c} + \lambda(\vec{b}\cdot\vec{c}) = 0.$ …

Common Mistakes

Mistake 1: Ignoring the zero $\hat{k}$-component of $\vec{c}$.

Why it's wrong: $\vec{c}=3\hat{i}+\hat{j}$ has a $\hat{k}$-component of $0$; forgetting it can misalign the dot-product terms. Correct approach: treat $\vec{c}=3\hat{i}+\hat{j}+0\hat{k}$ so every term is accounted for.

Mistake 2: Distributing $\lambda$ incorrectly. …