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JEE Mathematics 2024 Question with Solution

Let a⃗=3i^+j^−2k^\vec{a} = 3\hat{i} + \hat{j} - 2\hat{k}, b⃗=4i^+j^+7k^\vec{b} = 4\hat{i} + \hat{j} + 7\hat{k} and c⃗=i^−3j^+4k^\vec{c} = \hat{i} - 3\hat{j} + 4\hat{k} be three vectors. If a vector p⃗\vec{p} satisfies p⃗×b⃗=c⃗×b⃗\vec{p} \times \vec{b} = \vec{c} \times \vec{b} and p⃗⋅a⃗=0\vec{p} \cdot \vec{a} = 0, then p⃗⋅(i^−j^−k^)\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k}) is equal to:

  • A

    2424

  • B

    3636

  • C

    2828

  • D

    3232

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

  • a⃗=3i^+j^−2k^\vec{a} = 3\hat{i} + \hat{j} - 2\hat{k}
  • b⃗=4i^+j^+7k^\vec{b} = 4\hat{i} + \hat{j} + 7\hat{k}
  • c⃗=i^−3j^+4k^\vec{c} = \hat{i} - 3\hat{j} + 4\hat{k}
  • p⃗×b⃗=c⃗×b⃗\vec{p} \times \vec{b} = \vec{c} \times \vec{b}
  • p⃗⋅a⃗=0\vec{p} \cdot \vec{a} = 0

Find: p⃗⋅(i^−j^−k^)\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k})

From

p⃗×b⃗−c⃗×b⃗=0\vec{p} \times \vec{b} - \vec{c} \times \vec{b} = 0

we get

(p⃗−c⃗)×b⃗=0(\vec{p} - \vec{c}) \times \vec{b} = 0

Hence, p⃗−c⃗\vec{p} - \vec{c} is parallel to b⃗\vec{b}, so

p⃗−c⃗=λb⃗\vec{p} - \vec{c} = \lambda \vec{b}

that is,

p⃗=c⃗+λb⃗\vec{p} = \vec{c} + \lambda \vec{b}

Using the condition p⃗⋅a⃗=0\vec{p} \cdot \vec{a} = 0,

(c⃗+λb⃗)⋅a⃗=0(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 0 c⃗⋅a⃗+λ(b⃗⋅a⃗)=0\vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0

Now,

c⃗⋅a⃗=(1)(3)+(−3)(1)+(4)(−2)=3−3−8=−8\vec{c} \cdot \vec{a} = (1)(3) + (-3)(1) + (4)(-2) = 3 - 3 - 8 = -8

and

b⃗⋅a⃗=(4)(3)+(1)(1)+(7)(−2)=12+1−14=−1\vec{b} \cdot \vec{a} = (4)(3) + (1)(1) + (7)(-2) = 12 + 1 - 14 = -1

So,

−8+λ(−1)=0-8 + \lambda(-1) = 0 λ=−8\lambda = -8

Therefore,

p⃗=c⃗−8b⃗\vec{p} = \vec{c} - 8\vec{b} p⃗=(i^−3j^+4k^)−8(4i^+j^+7k^)\vec{p} = (\hat{i} - 3\hat{j} + 4\hat{k}) - 8(4\hat{i} + \hat{j} + 7\hat{k}) p⃗=−31i^−11j^−52k^\vec{p} = -31\hat{i} - 11\hat{j} - 52\hat{k}

Now compute

p⃗⋅(i^−j^−k^)\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k}) =(−31)(1)+(−11)(−1)+(−52)(−1)= (-31)(1) + (-11)(-1) + (-52)(-1) =−31+11+52=32= -31 + 11 + 52 = 32

Therefore, the correct option is D, and the required value is 3232.

Common mistakes

  • Assuming from (p⃗−c⃗)×b⃗=0(\vec{p} - \vec{c}) \times \vec{b} = 0 that p⃗−c⃗=0\vec{p} - \vec{c} = 0. This is wrong because zero cross product only implies the vectors are parallel. Instead, write p⃗−c⃗=λb⃗\vec{p} - \vec{c} = \lambda \vec{b}.

  • Making an error in the dot products c⃗⋅a⃗\vec{c} \cdot \vec{a} or b⃗⋅a⃗\vec{b} \cdot \vec{a}, especially with the negative signs. This changes λ\lambda and gives a wrong vector p⃗\vec{p}. Compute each component product carefully before adding.

  • Using the wrong signs while evaluating p⃗⋅(i^−j^−k^)\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k}). The coefficients of the second and third components are −1-1, so the expression is x−y−zx - y - z, not x+y+zx + y + z.

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