Maths Q15 – JEE-MAIN 2026

Let

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

and

\vec{b} = 6 \hat{i} + 3 \hat{j} + 3 \hat{k}

. Then the square of the area of the triangle with adjacent sides determined by the vectors

(2 \vec{a} + 3 \vec{b})

and

(\vec{a} - \vec{b})

is:

Clue 1 of 3: Vector Operations

First, calculate the resultant vectors that form the adjacent sides of the triangle.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Calculate the adjacent side vectors✦ Active

Given $\vec{a} = 2 \hat{i} + 3 \hat{j} + 3 \hat{k}$ and $\vec{b} = 6 \hat{i} + 3 \hat{j} + 3 \hat{k}$ . Let the adjacent sides of the triangle be $\vec{u} = 2 \vec{a} + 3 \vec{b}$ and $\vec{v} = \vec{a} - \vec{b}$ .

\vec{u} = 2 (2 \hat{i} + 3 \hat{j} + 3 \hat{k}) + 3 (6 \hat{i} + 3 \hat{j} + 3 \hat{k}) = (4 \hat{i} + 6 \hat{j} + 6 \hat{k}) + (18 \hat{i} + 9 \hat{j} + 9 \hat{k}) = 22 \hat{i} + 15 \hat{j} + 15 \hat{k}

\vec{v} = (2 \hat{i} + 3 \hat{j} + 3 \hat{k}) - (6 \hat{i} + 3 \hat{j} + 3 \hat{k}) = (2 - 6) \hat{i} + (3 - 3) \hat{j} + (3 - 3) \hat{k} = - 4 \hat{i}

Step 2: Compute the cross product of the side vectors○ Expand

The cross product of $\vec{u}$ and $\vec{v}$ is calculated as follows:

\vec{u} \times \vec{v} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 22 & 15 & 15 \\ - 4 & 0 & 0 \end{matrix} |

= \hat{i} (15 \times 0 - 15 \times 0) - \hat{j} (22 \times 0 - 15 \times (- 4)) + \hat{k} (22 \times 0 - 15 \times (- 4))

= 0 \hat{i} - (- 60) \hat{j} + (- 60) \hat{k} = - 60 \hat{j} + 60 \hat{k}

💡 Teacher's Secret Hint

Be careful with the signs when expanding the determinant for the cross product.

Step 3: Calculate the square of the area of the triangle○ Expand

The magnitude of the cross product is:

| \vec{u} \times \vec{v} | = \sqrt{(- 60)^{2} + (60)^{2}} = \sqrt{3600 + 3600} = \sqrt{2 \times 3600} = 60 \sqrt{2}

The area of the triangle is $\frac{1}{2} | \vec{u} \times \vec{v} |$ . The square of the area is:

{(\frac{1}{2} | \vec{u} \times \vec{v} |)}^{2} = {(\frac{1}{2} \times 60 \sqrt{2})}^{2} = (30 \sqrt{2})^{2} = 30^{2} \times (\sqrt{2})^{2} = 900 \times 2 = 1800

💡 Teacher's Secret Hint

Remember to square the entire expression for the area, including the $\frac{1}{2}$ factor.

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

Maths Question 15 – JEE-MAIN 2026

Clue 1 of 3: Vector Operations

🧠 Full Solution Path