JEE Mathematics 2026 Question with Solution

Given: $\vec{a} = (1,-2,3)$, $\vec{b} = (2,1,-1)$, $\vec{v} = \vec{a} \times \vec{b}$, and $\vec{v} \cdot \vec{c} = 11$ where $\vec{c} = (\lambda,1,1)$.

Find: $9p^2$, where $p$ is the projection of $\vec{b}$ on $\vec{c}$.

Using the cross product,

$$\vec{v} = \vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{vmatrix} = (-1,7,5)$$

Now,

$$\vec{v} \cdot \vec{c} = (-1,7,5) \cdot (\lambda,1,1) = -\lambda + 7 + 5 = 12 - \lambda$$

Given $\vec{v} \cdot \vec{c} = 11$, so

$$12 - \lambda = 11$$ $$\lambda = 1$$

Hence,

$$\vec{c} = (1,1,1), \qquad |\vec{c}| = \sqrt{3}$$

The projection of $\vec{b}$ on $\vec{c}$ is

$$p = \frac{\vec{b} \cdot \vec{c}}{|\vec{c}|}$$

So,

$$p = \frac{2(1) + 1(1) - 1(1)}{\sqrt{3}} = \frac{2}{\sqrt{3}}$$

Therefore,

$$9p^2 = 9 \left(\frac{2}{\sqrt{3}}\right)^2 = 9 \cdot \frac{4}{3} = 12$$

Therefore, the correct option is A.

Given: $\vec{v} = \vec{a} \times \vec{b}$ and $\vec{v} \cdot \vec{c} = 11$.

Find: the value of $9p^2$.

The hint uses the scalar triple product:

$$[\vec{a}\ \vec{b}\ \vec{c}] = (\vec{a} \times \vec{b}) \cdot \vec{c}$$

From the working,

$$\vec{a} = (1,-2,3), \qquad \vec{b} = (2,1,-1), \qquad \vec{c} = (\lambda,1,1)$$

First compute

$$\vec{a} \times \vec{b} = (-1,7,5)$$

Then use

$$(-1,7,5) \cdot (\lambda,1,1) = 11$$

which gives

$$-\lambda + 7 + 5 = 11$$ $$12 - \lambda = 11$$ $$\lambda = 1$$

So,

$$\vec{c} = (1,1,1)$$

Now compute projection of $\vec{b}$ on $\vec{c}$:

$$p = \frac{\vec{b} \cdot \vec{c}}{|\vec{c}|}$$

Here,

$$\vec{b} \cdot \vec{c} = (2,1,-1) \cdot (1,1,1) = 2 + 1 - 1 = 2$$

and

$$|\vec{c}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}$$

Thus,

$$p = \frac{2}{\sqrt{3}}$$

Hence,

$$9p^2 = 9 \cdot \frac{4}{3} = 12$$

Therefore, the value of $9p^2$ is $12$.