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JEE Mathematics 2024 Question with Solution

Let a⃗=i^+2j^+3k^\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, b⃗=2i^+3j^−5k^\vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}, and c⃗=3i^−j^+λk^\vec{c} = 3\hat{i} - \hat{j} + \lambda \hat{k} be three vectors. Let r⃗\vec{r} be a unit vector along b⃗+c⃗\vec{b} + \vec{c}. If r⃗⋅a⃗=3\vec{r} \cdot \vec{a} = 3, then 3λ3\lambda is equal to:

  • A

    2727

  • B

    2525

  • C

    2525

  • D

    2121

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: a⃗=i^+2j^+3k^\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, b⃗=2i^+3j^−5k^\vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}, c⃗=3i^−j^+λk^\vec{c} = 3\hat{i} - \hat{j} + \lambda \hat{k}, and r⃗\vec{r} is a unit vector along b⃗+c⃗\vec{b} + \vec{c} with r⃗⋅a⃗=3\vec{r} \cdot \vec{a} = 3.

Find: 3λ3\lambda.

First compute b⃗+c⃗\vec{b} + \vec{c}:

b⃗+c⃗=(2+3)i^+(3−1)j^+(−5+λ)k^=5i^+2j^+(λ−5)k^\vec{b} + \vec{c} = (2+3)\hat{i} + (3-1)\hat{j} + (-5+\lambda)\hat{k} = 5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}

Its magnitude is:

∣b⃗+c⃗∣=52+22+(λ−5)2=29+(λ−5)2|\vec{b} + \vec{c}| = \sqrt{5^2 + 2^2 + (\lambda - 5)^2} = \sqrt{29 + (\lambda - 5)^2}

Therefore the unit vector along b⃗+c⃗\vec{b} + \vec{c} is:

r⃗=5i^+2j^+(λ−5)k^29+(λ−5)2\vec{r} = \frac{5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}}{\sqrt{29 + (\lambda - 5)^2}}

Now use the condition r⃗⋅a⃗=3\vec{r} \cdot \vec{a} = 3:

(5i^+2j^+(λ−5)k^)⋅(i^+2j^+3k^)29+(λ−5)2=3\frac{(5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k})}{\sqrt{29 + (\lambda - 5)^2}} = 3

Evaluate the dot product:

5(1)+2(2)+(λ−5)(3)=5+4+3λ−15=3λ−65(1) + 2(2) + (\lambda - 5)(3) = 5 + 4 + 3\lambda - 15 = 3\lambda - 6

So,

3λ−629+(λ−5)2=3\frac{3\lambda - 6}{\sqrt{29 + (\lambda - 5)^2}} = 3

Cross-multiplying and simplifying:

λ−2=29+(λ−5)2\lambda - 2 = \sqrt{29 + (\lambda - 5)^2}

Squaring both sides:

(λ−2)2=29+(λ−5)2(\lambda - 2)^2 = 29 + (\lambda - 5)^2

Expand both sides:

λ2−4λ+4=29+λ2−10λ+25\lambda^2 - 4\lambda + 4 = 29 + \lambda^2 - 10\lambda + 25 −4λ+4=54−10λ-4\lambda + 4 = 54 - 10\lambda 6λ=506\lambda = 50 λ=253\lambda = \frac{25}{3}

Hence,

3λ=253\lambda = 25

Therefore, the correct option is B.

Direct Equation Simplification

Given: r⃗\vec{r} is the unit vector along b⃗+c⃗=5i^+2j^+(λ−5)k^\vec{b} + \vec{c} = 5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k} and r⃗⋅a⃗=3\vec{r} \cdot \vec{a} = 3.

Find: 3λ3\lambda.

Write immediately:

r⃗=5i^+2j^+(λ−5)k^29+(λ−5)2\vec{r} = \frac{5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}}{\sqrt{29 + (\lambda - 5)^2}}

Then

r⃗⋅a⃗=5+4+3(λ−5)29+(λ−5)2=3\vec{r} \cdot \vec{a} = \frac{5 + 4 + 3(\lambda - 5)}{\sqrt{29 + (\lambda - 5)^2}} = 3

So,

3λ−629+(λ−5)2=3\frac{3\lambda - 6}{\sqrt{29 + (\lambda - 5)^2}} = 3 λ−2=29+(λ−5)2\lambda - 2 = \sqrt{29 + (\lambda - 5)^2} (λ−2)2=29+(λ−5)2(\lambda - 2)^2 = 29 + (\lambda - 5)^2

Cancel λ2\lambda^2 after expansion and solve the resulting linear equation:

λ2−4λ+4=λ2−10λ+54\lambda^2 - 4\lambda + 4 = \lambda^2 - 10\lambda + 54 6λ=506\lambda = 50 λ=253\lambda = \frac{25}{3}

Hence 3λ=253\lambda = 25. The correct option is B.

Common mistakes

  • Students often take the unit vector along b⃗+c⃗\vec{b} + \vec{c} as just b⃗+c⃗\vec{b} + \vec{c}. That is incorrect because a unit vector must have magnitude 11. Always divide by ∣b⃗+c⃗∣|\vec{b} + \vec{c}| first.

  • A common error is computing b⃗+c⃗\vec{b} + \vec{c} incorrectly, especially the j^\hat{j} and k^\hat{k} components. Here 3+(−1)=23 + (-1) = 2 and −5+λ=λ−5-5 + \lambda = \lambda - 5. Add components carefully.

  • Many students make a mistake in the dot product by writing 5+2+(λ−5)5 + 2 + (\lambda - 5) or by forgetting to multiply the k^\hat{k} components by 33. Use component-wise multiplication: 5⋅1+2⋅2+(λ−5)⋅35\cdot1 + 2\cdot2 + (\lambda - 5)\cdot3.

  • After squaring, students may stop at λ=253\lambda = \frac{25}{3} and forget that the question asks for 3λ3\lambda. Always check the final quantity required before marking the answer.

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