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MathMedMain 2026 • 2nd Apr - Shift 2

Find Magnitude Squared of Linear Combination of Two Vectors

Let the vectors a⃗=−i^+j^+3k^\vec{a} = -\hat{i} + \hat{j} + 3\hat{k} and b⃗=i^+3j^+k^\vec{b} = \hat{i} + 3\hat{j} + \hat{k}. For some λ,μ∈R\lambda, \mu \in \mathbb{R}, let c⃗=λa⃗+μb⃗\vec{c} = \lambda\vec{a} + \mu\vec{b}. If c⃗⋅(3i^−6j^+2k^)=10\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10 and c⃗⋅(i^+j^+k^)=−2\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2, then ∣c⃗∣2|\vec{c}|^2 is equal to :

Options

A

8

B

12

C

14

Correct
D

15

Topics & Concepts

Vector AlgebraVector AdditionDot and Cross Product

Step-by-Step Solution

To find the value of ∣c⃗∣2|\vec{c}|^2, we start by expressing the vector c⃗\vec{c} in component form in terms of the real scalars λ\lambda and μ\mu.

Given: a⃗=−i^+j^+3k^\vec{a} = -\hat{i} + \hat{j} + 3\hat{k} b⃗=i^+3j^+k^\vec{b} = \hat{i} + 3\hat{j} + \hat{k}

Since c⃗=λa⃗+μb⃗\vec{c} = \lambda\vec{a} + \mu\vec{b}, we have: c⃗=λ(−i^+j^+3k^)+μ(i^+3j^+k^)\vec{c} = \lambda(-\hat{i} + \hat{j} + 3\hat{k}) + \mu(\hat{i} + 3\hat{j} + \hat{k}) c⃗=(μ−λ)i^+(λ+3μ)j^+(3λ+μ)k^\vec{c} = (\mu - \lambda)\hat{i} + (\lambda + 3\mu)\hat{j} + (3\lambda + \mu)\hat{k}

We are given two dot product conditions:

Step 1: Using the first condition

c⃗⋅(3i^−6j^+2k^)=10\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10

Taking the dot product: λ(a⃗⋅(3i^−6j^+2k^))+μ(b⃗⋅(3i^−6j^+2k^))=10\lambda \left( \vec{a} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) \right) + \mu \left( \vec{b} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) \right) = 10

Calculating the individual dot products: a⃗⋅(3i^−6j^+2k^)=(−1)(3)+(1)(−6)+(3)(2)=−3−6+6=−3\vec{a} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = (-1)(3) + (1)(-6) + (3)(2) = -3 - 6 + 6 = -3 b⃗⋅(3i^−6j^+2k^)=(1)(3)+(3)(−6)+(1)(2)=3−18+2=−13\vec{b} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = (1)(3) + (3)(-6) + (1)(2) = 3 - 18 + 2 = -13

Substituting these values back: −3λ−13μ=10— (Equation 1)-3\lambda - 13\mu = 10 \quad \text{--- (Equation 1)}

Step 2: Using the second condition

c⃗⋅(i^+j^+k^)=−2\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2

Taking the dot product: λ(a⃗⋅(i^+j^+k^))+μ(b⃗⋅(i^+j^+k^))=−2\lambda \left( \vec{a} \cdot (\hat{i} + \hat{j} + \hat{k}) \right) + \mu \left( \vec{b} \cdot (\hat{i} + \hat{j} + \hat{k}) \right) = -2

Calculating the individual dot products: a⃗⋅(i^+j^+k^)=−1+1+3=3\vec{a} \cdot (\hat{i} + \hat{j} + \hat{k}) = -1 + 1 + 3 = 3 b⃗⋅(i^+j^+k^)=1+3+1=5\vec{b} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1 + 3 + 1 = 5

Substituting these values back: 3λ+5μ=−2— (Equation 2)3\lambda + 5\mu = -2 \quad \text{--- (Equation 2)}

Step 3: Solving for λ\lambda and μ\mu

Adding Equation 1 and Equation 2: (−3λ−13μ)+(3λ+5μ)=10+(−2)(-3\lambda - 13\mu) + (3\lambda + 5\mu) = 10 + (-2) −8μ=8  ⟹  μ=−1-8\mu = 8 \implies \mu = -1

Substitute μ=−1\mu = -1 into Equation 2: 3λ+5(−1)=−23\lambda + 5(-1) = -2 3λ−5=−2  ⟹  3λ=3  ⟹  λ=13\lambda - 5 = -2 \implies 3\lambda = 3 \implies \lambda = 1

Step 4: Finding c⃗\vec{c} and ∣c⃗∣2|\vec{c}|^2

Substitute λ=1\lambda = 1 and μ=−1\mu = -1 into the expression for c⃗\vec{c}: c⃗=1a⃗−1b⃗=a⃗−b⃗\vec{c} = 1\vec{a} - 1\vec{b} = \vec{a} - \vec{b} c⃗=(−i^+j^+3k^)−(i^+3j^+k^)\vec{c} = (-\hat{i} + \hat{j} + 3\hat{k}) - (\hat{i} + 3\hat{j} + \hat{k}) c⃗=−2i^−2j^+2k^\vec{c} = -2\hat{i} - 2\hat{j} + 2\hat{k}

Now, compute the magnitude squared ∣c⃗∣2|\vec{c}|^2: ∣c⃗∣2=(−2)2+(−2)2+(2)2=4+4+4=12|\vec{c}|^2 = (-2)^2 + (-2)^2 + (2)^2 = 4 + 4 + 4 = 12

Thus, ∣c⃗∣2|\vec{c}|^2 is equal to 1212.

Find Magnitude Squared of Linear Combination of Two Vectors | Mathematics PYQ Solution - JEE Challenger