What is the formula for the cross product of two vectors?

The cross product a x b is calculated using the determinant of a 3x3 matrix with unit vectors i, j, k in the first row.

How is the scalar triple product related to this problem?

The condition v . c = 11 is equivalent to the scalar triple product [a b c] = 11.

What is the formula for the projection of vector b on vector c?

The length of the projection p is given by |b . c| / |c|.

How do we find the value of λ?

By expanding the determinant for the scalar triple product [a b c] and equating it to 11, we find λ = 2.

What is the magnitude of vector c when λ = 2?

|c| = sqrt(2^2 + 1^2 + 1^2) = sqrt(6).

What is the value of the dot product b . c?

With λ = 2, b . c = (2)(2) + (1)(1) + (-1)(1) = 4 + 1 - 1 = 4.

p = (b . c) / |c| = 4 / sqrt(6).

What is the final step to find 9p^2?

9p^2 = 9 * (16 / 6) = 3 * 8 = 24. Wait, re-calculating... 9 * (16/6) = 3 * 8 = 24. Re-check: 9 * 8/3 = 24.

Is there a mistake in the options?

Based on calculations, 9p^2 is 24. If the options show 12, check the dot product or λ value again.

What is the projection of a vector?

It is the scalar component of one vector along the direction of another vector.

Let a = i - 2j + 3k, b = 2i + j - k, c = λi + j + k and v = a x b. If v . c = 11 and the length of the projection of b on c is p, then 9p^2 is equal to

Let a = i – 2j + 3k, b = 2i + j – k, c = λi + j + k and v = a x b. If v . c = 11 and the length of the projection of b on c is p, then 9p^2 is equal to | JEE Main Mathematics

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MathJEE 2026

QVectors

Let $\vec{a} = \hat{i} – 2\hat{j} + 3\hat{k}$, $\vec{b} = 2\hat{i} + \hat{j} – \hat{k}$, $\vec{c} = \lambda\hat{i} + \hat{j} + \hat{k}$ and $\vec{v} = \vec{a} \times \vec{b}$. If $\vec{v} \cdot \vec{c} = 11$ and the length of the projection of $\vec{b}$ on $\vec{c}$ is $p$, then $9p^2$ is equal to :

A) $9$ B) $6$ C) $4$ D) $12$

✅ Correct Answer

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Solution Steps

Calculate vector $\vec{v} = \vec{a} \times \vec{b}$

$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{vmatrix} = \hat{i}(2-3) – \hat{j}(-1-6) + \hat{k}(1+4) = -\hat{i} + 7\hat{j} + 5\hat{k}$

Use condition $\vec{v} \cdot \vec{c} = 11$ to find $\lambda$

$(-\hat{i} + 7\hat{j} + 5\hat{k}) \cdot (\lambda\hat{i} + \hat{j} + \hat{k}) = 11$

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

Define vector $\vec{c}$ and find its magnitude

Since $\lambda = 1$, $\vec{c} = \hat{i} + \hat{j} + \hat{k}$.

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

Calculate length of projection $p$

Projection of $\vec{b}$ on $\vec{c}$ is $p = \frac{|\vec{b} \cdot \vec{c}|}{|\vec{c}|}$.

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

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

Calculate $9p^2$

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

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

$9p^2 = 12$ (Option D)

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Theory

1. Vector Cross Product

Cross product do vectors ka product hota hai jo ek naya vector produce karta hai jo original vectors ke perpendicular hota hai. Isse nikalne ke liye determinant method ka use kiya jata hai. Is question mein $\vec{v} = \vec{a} \times \vec{b}$ nikalne ke liye humne determinant simplify kiya. Ye physics aur geometry mein rotation aur area nikalne ke kaam aata hai.

2. Scalar Triple Product (STP)

Condition $\vec{v} \cdot \vec{c} = 11$ actually $(\vec{a} \times \vec{b}) \cdot \vec{c}$ hai, jise Scalar Triple Product $[\vec{a} \vec{b} \vec{c}]$ kehte hain. Ye ek parallelepiped ka volume represent karta hai. STP tabhi useful hota hai jab humein teen vectors ki coplanarity check karni ho ya koi unknown scalar (jaise $\lambda$) find karna ho.

3. Projection of a Vector

Projection ka matlab hota hai ek vector ka ‘parchaai’ dusre vector ki direction mein. Length of projection $p$ formula $\frac{|\vec{b} \cdot \vec{c}|}{|\vec{c}|}$ se di jati hai. Isme numerator dot product hota hai aur denominator target vector ka magnitude. Ye component analysis mein bahut important role play karta hai.

4. Dot Product Properties

Dot product ek scalar quantity hai jo $x_1x_2 + y_1y_2 + z_1z_2$ se calculate hoti hai. Agar do vectors perpendicular honge toh dot product 0 hoga. JEE Main problems mein unknown values nikalne ke liye equations dot product se hi banti hain. Yahan $\vec{v} \cdot \vec{c}$ ki value se humne $\lambda=1$ determine kiya.

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FAQs

Projection vector aur length of projection mein kya antar hai?

Length of projection ek scalar (value) hai, jabki projection vector $(\frac{\vec{b} \cdot \vec{c}}{|\vec{c}|^2})\vec{c}$ ek vector hota hai.

Scalar triple product zero kab hota hai?

Jab teeno vectors coplanar hote hain, tab STP zero ho jata hai.

Unit vector along c kaise nikalenge?

Vector $\vec{c}$ ko uske magnitude $|\vec{c}|$ se divide karke, yani $\hat{c} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.

Determinant mein rows interchange karne par kya hota hai?

Ek baar interchange karne par STP ka sign badal jata hai. Do baar karne par sign same rehta hai.

Projection negative ho sakti hai?

Dot product negative ho sakta hai (agar angle obtuse ho), par ‘length’ of projection hamesha positive value (magnitude) li jati hai.

λ nikalne ke liye determinant kyun use nahi kiya?

Aap direct determinant $\begin{vmatrix} 1 & -2 & 3 \\ 2 & 1 & -1 \\ \lambda & 1 & 1 \end{vmatrix} = 11$ bhi solve kar sakte the, result $\lambda = 1$ hi aata.

9p^2 nikalne ka shortcut?

Shortcut nahi, par precision zaroori hai. $p = 2/\sqrt{3}$ ke baad square karke multiply karna sabse safe hai.

v vector ka direction kya hai?

$\vec{v}$ hamesha $\vec{a}$ aur $\vec{b}$ dono ke perpendicular hota hai (Right hand thumb rule).

b on c aur c on b same hota hai?

Nahi, kyunki denominator change ho jayega ($|\vec{c}|$ vs $|\vec{b}|$).

Cross product commutative hota hai?

Nahi, $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$. Isliye order important hai.

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