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Maths Question 14 – JEE-MAIN 2026

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Let the vectors a→=−i^+j^+3k^ and b→=i^+3j^+k^. For some λ,μ∈R, let c→=λa→+μb→. If c→⋅(3i^−6j^+2k^)=10 and c→⋅(i^+j^+k^)=−2, then |c→|2 is equal to :

Express the vector c→ in terms of its components using the given linear combination of vectors a→ and b→.

Step 1: Express c→ in terms of λ and μ and set up equations✦ Active

Given a→=−i^+j^+3k^ and b→=i^+3j^+k^, we have c→=λa→+μb→. Expanding this, we get:

c→=(−λ+μ)i^+(λ+3μ)j^+(3λ+μ)k^

Using the given dot product conditions, we form two linear equations:

c→⋅(3i^−6j^+2k^)=10⟹3(−λ+μ)−6(λ+3μ)+2(3λ+μ)=10⟹−3λ−13μ=10(1)
c→⋅(i^+j^+k^)=−2⟹(−λ+μ)+(λ+3μ)+(3λ+μ)=−2⟹3λ+5μ=−2(2)
Step 2: Solve for λ and μ○ Expand

We have the system of equations:

1)−3λ−13μ=10
2)3λ+5μ=−2

Adding equation (1) and (2) eliminates λ:

(−3λ−13μ)+(3λ+5μ)=10+(−2)⟹−8μ=8⟹μ=−1

Substitute μ=−1 into equation (2):

3λ+5(−1)=−2⟹3λ−5=−2⟹3λ=3⟹λ=1
Step 3: Calculate c→ and its squared magnitude○ Expand

Substitute λ=1 and μ=−1 back into the expression for c→:

c→=(−1+(−1))i^+(1+3(−1))j^+(3(1)+(−1))k^
c→=−2i^−2j^+2k^

Finally, calculate the squared magnitude of c→:

|c→|2=(−2)2+(−2)2+(2)2=4+4+4=12
💡 Teacher's Secret Hint

Double-check your arithmetic when solving the system of equations and calculating the magnitude.

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