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MCQEasyJEE 2025Basics of Vectors

JEE Mathematics 2025 Question with Solution

If the components of a⃗=αi^+βj^+γk^\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k} along and perpendicular to b⃗=3i^+j^−k^\vec{b} = 3\hat{i} + \hat{j} - \hat{k} respectively are 1611(3i^+j^−k^)\frac{16}{11} (3\hat{i} + \hat{j} - \hat{k}) and 111(−4i^−5j^−17k^)\frac{1}{11} (-4\hat{i} - 5\hat{j} - 17\hat{k}), then α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2 is equal to:

  • A

    1818

  • B

    2626

  • C

    2323

  • D

    1616

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The components of a⃗\vec{a} along and perpendicular to b⃗=3i^+j^−k^\vec{b} = 3\hat{i} + \hat{j} - \hat{k} are

a⃗∥=1611(3i^+j^−k^)\vec{a}_{\parallel} = \frac{16}{11}(3\hat{i} + \hat{j} - \hat{k})

and

a⃗⊥=111(−4i^−5j^−17k^)\vec{a}_{\perp} = \frac{1}{11}(-4\hat{i} - 5\hat{j} - 17\hat{k})

Find: α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2 for a⃗=αi^+βj^+γk^\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}.

Since a vector equals the sum of its parallel and perpendicular components,

a⃗=a⃗∥+a⃗⊥\vec{a} = \vec{a}_{\parallel} + \vec{a}_{\perp}

So,

a⃗=1611(3i^+j^−k^)+111(−4i^−5j^−17k^)\vec{a} = \frac{16}{11}(3\hat{i} + \hat{j} - \hat{k}) + \frac{1}{11}(-4\hat{i} - 5\hat{j} - 17\hat{k})

Combining components,

a⃗=111[(16×3−4)i^+(16−5)j^+(−16−17)k^]\vec{a} = \frac{1}{11}\left[(16 \times 3 - 4)\hat{i} + (16 - 5)\hat{j} + (-16 - 17)\hat{k}\right] a⃗=111(44i^+11j^−33k^)\vec{a} = \frac{1}{11}(44\hat{i} + 11\hat{j} - 33\hat{k}) a⃗=4i^+j^−3k^\vec{a} = 4\hat{i} + \hat{j} - 3\hat{k}

Therefore,

α=4,β=1,γ=−3\alpha = 4, \quad \beta = 1, \quad \gamma = -3

Now,

α2+β2+γ2=42+12+(−3)2\alpha^2 + \beta^2 + \gamma^2 = 4^2 + 1^2 + (-3)^2 =16+1+9=26= 16 + 1 + 9 = 26

Therefore, the correct option is B and the value of α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2 is 2626.

Component-wise Expansion

Given:

a⃗∥=1611(3i^+j^−k^),a⃗⊥=111(−4i^−5j^−17k^)\vec{a}_{\parallel} = \frac{16}{11}(3\hat{i} + \hat{j} - \hat{k}), \qquad \vec{a}_{\perp} = \frac{1}{11}(-4\hat{i} - 5\hat{j} - 17\hat{k})

Find: α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2.

Write the total vector as

a⃗=a⃗∥+a⃗⊥\vec{a} = \vec{a}_{\parallel} + \vec{a}_{\perp}

Expand each part:

a⃗∥=4811i^+1611j^−1611k^\vec{a}_{\parallel} = \frac{48}{11}\hat{i} + \frac{16}{11}\hat{j} - \frac{16}{11}\hat{k} a⃗⊥=−411i^−511j^−1711k^\vec{a}_{\perp} = -\frac{4}{11}\hat{i} - \frac{5}{11}\hat{j} - \frac{17}{11}\hat{k}

Add corresponding components:

α=4811−411=4411=4\alpha = \frac{48}{11} - \frac{4}{11} = \frac{44}{11} = 4 β=1611−511=1111=1\beta = \frac{16}{11} - \frac{5}{11} = \frac{11}{11} = 1 γ=−1611−1711=−3311=−3\gamma = -\frac{16}{11} - \frac{17}{11} = -\frac{33}{11} = -3

Hence,

α2+β2+γ2=42+12+(−3)2=26\alpha^2 + \beta^2 + \gamma^2 = 4^2 + 1^2 + (-3)^2 = 26

Therefore, the required value is 2626.

Common mistakes

  • Adding only the parallel component and ignoring the perpendicular component. This is wrong because the vector a⃗\vec{a} is the sum of both components. Always use a⃗=a⃗∥+a⃗⊥\vec{a} = \vec{a}_{\parallel} + \vec{a}_{\perp}.

  • Making a sign error in the k^\hat{k} component. The terms are −1611k^-\frac{16}{11}\hat{k} and −1711k^-\frac{17}{11}\hat{k}, so they add to −3311k^=−3k^-\frac{33}{11}\hat{k} = -3\hat{k}, not a positive value.

  • Finding α,β,γ\alpha, \beta, \gamma correctly but then computing α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2 incorrectly. After obtaining 4,1,−34, 1, -3, square each component carefully: 16,1,916, 1, 9.

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