Let a = i + 2j + k and b = 2i + j - k. Let c be a unit vector in the plane of the vectors a and b and be perpendicular to a. Then such a vector c is :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q14 – JEE-MAIN 2025

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Home Maths Vector AlgebraQuestion #524

Let

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

and

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

. Let

\hat{c}

be a unit vector in the plane of the vectors

\vec{a}

and

\vec{b}

and be perpendicular to

\vec{a}

. Then such a vector

\hat{c}

is :

Clue 1 of 3: Vector Properties

A vector in the plane of two other vectors can be expressed as their linear combination. The dot product of two perpendicular vectors is zero.

🥷

Ninja Strategy•Coplanarity Check

First, calculate the normal vector to the plane formed by $\vec{a}$ and $\vec{b}$ as $\vec{n} = \vec{a} \times \vec{b}$ . Then, check which option $\hat{c}$ satisfies the coplanarity condition $\hat{c} \cdot \vec{n} = 0$ .

🧠 Full Solution Path

Step 1: Express

\hat{c}

as a linear combination and apply perpendicularity✦ Active

Given $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - \hat{k}$ . Since $\hat{c}$ is in the plane of $\vec{a}$ and $\vec{b}$ , we can write $\hat{c} = x \vec{a} + y \vec{b}$ for some scalars $x, y$ . Substituting the given vectors:

\hat{c} = x (\hat{i} + 2 \hat{j} + \hat{k}) + y (2 \hat{i} + \hat{j} - \hat{k}) = (x + 2 y) \hat{i} + (2 x + y) \hat{j} + (x - y) \hat{k}

Since $\hat{c}$ is perpendicular to $\vec{a}$ , their dot product is zero: $\hat{c} \cdot \vec{a} = 0$ .

(x + 2 y) (1) + (2 x + y) (2) + (x - y) (1) = 0

Simplifying the equation:

x + 2 y + 4 x + 2 y + x - y = 0 ⟹ 6 x + 3 y = 0 ⟹ y = - 2 x

Step 2: Substitute the relationship and simplify

\hat{c}

○ Expand

Substitute $y = - 2 x$ back into the expression for $\hat{c}$ :

\hat{c} = (x + 2 (- 2 x)) \hat{i} + (2 x + (- 2 x)) \hat{j} + (x - (- 2 x)) \hat{k}

This simplifies to:

\hat{c} = (x - 4 x) \hat{i} + (2 x - 2 x) \hat{j} + (x + 2 x) \hat{k} = - 3 x \hat{i} + 0 \hat{j} + 3 x \hat{k} = 3 x (- \hat{i} + \hat{k})

Step 3: Normalize the vector to find the unit vector○ Expand

Since $\hat{c}$ is a unit vector, its magnitude must be 1. We find the magnitude of $3 x (- \hat{i} + \hat{k})$ :

| \hat{c} | = | 3 x | \sqrt{(- 1)^{2} + 0^{2} + 1^{2}} = | 3 x | \sqrt{1 + 1} = | 3 x | \sqrt{2}

Setting the magnitude to 1:

| 3 x | \sqrt{2} = 1 ⟹ | x | = \frac{1}{3 \sqrt{2}}

Thus, $x = \pm \frac{1}{3 \sqrt{2}}$ . Substituting this back into the expression for $\hat{c}$ :

\hat{c} = 3 (\pm \frac{1}{3 \sqrt{2}}) (- \hat{i} + \hat{k}) = \pm \frac{1}{\sqrt{2}} (- \hat{i} + \hat{k})

Comparing this with the given options, option 3 matches one of the possible unit vectors.

💡 Teacher's Secret Hint

Remember that a unit vector can have two opposite directions.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 14 – JEE-MAIN 2025

Clue 1 of 3: Vector Properties

🧠 Full Solution Path