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Vector Algebra question

2025 · 2 Apr · Shift 2 · Q44
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  5. /2025 · 2 Apr · Shift 2 · Q44

Vector Algebra question

2025 · 2 Apr · Shift 2 · Q44

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a→=2i^−3j^+k^, b→=3i^+2j^+5k^\overrightarrow{\mathrm{a}}=2 \hat{i}-3 \hat{j}+\hat{k}, \quad \overrightarrow{\mathrm{~b}}=3 \hat{i}+2 \hat{j}+5 \hat{k}a=2i^−3j^​+k^, b=3i^+2j^​+5k^ and a vector c→\overrightarrow{\mathrm{c}}c be such that (a⃗−c⃗)×b⃗=−18i^−3j^+12k^(\vec{a}-\vec{c}) \times \vec{b}=-18 \hat{i}-3 \hat{j}+12 \hat{k}(a−c)×b=−18i^−3j^​+12k^ and a⃗⋅c⃗=3\vec{a} \cdot \vec{c}=3a⋅c=3. If b⃗×c⃗=d⃗\vec{b} \times \vec{c}=\vec{d}b×c=d, then ∣a⃗⋅d⃗∣|\vec{a} \cdot \vec{d}|∣a⋅d∣ is equal to :
  1. A
    15
  2. B
    18
  3. C
    12
  4. D
    9
View written solutionFree

Correct answer: A

  1. Given vectors
a⃗=(2,−3,1),b⃗=(3,2,5),c⃗=(x,y,z)\vec a=(2,-3,1),\qquad \vec b=(3,2,5),\qquad \vec c=(x,y,z)a=(2,−3,1),b=(3,2,5),c=(x,y,z)

Also,

(a⃗−c⃗)×b⃗=−18i^−3j^+12k^(\vec a-\vec c)\times \vec b=-18\hat i-3\hat j+12\hat k(a−c)×b=−18i^−3j^​+12k^

and

a⃗⋅c⃗=3.\vec a\cdot \vec c=3.a⋅c=3.

We need to find

d⃗=b⃗×c⃗,\vec d=\vec b\times \vec c,d=b×c,

then compute

∣a⃗⋅d⃗∣.|\vec a\cdot \vec d|.∣a⋅d∣.
  1. Use the cross product condition

Since cross product is distributive,

(a⃗−c⃗)×b⃗=a⃗×b⃗−c⃗×b⃗.(\vec a-\vec c)\times \vec b=\vec a\times \vec b-\vec c\times \vec b.(a−c)×b=a×b−c×b.

But

d⃗=b⃗×c⃗  ⟹  c⃗×b⃗=−(b⃗×c⃗)=−d⃗.\vec d=\vec b\times \vec c \implies \vec c\times \vec b=-(\vec b\times \vec c)=-\vec d.d=b×c⟹c×b=−(b×c)=−d.

Hence,

(a⃗−c⃗)×b⃗=a⃗×b⃗+d⃗.(\vec a-\vec c)\times \vec b=\vec a\times \vec b+\vec d.(a−c)×b=a×b+d.

So first compute a⃗×b⃗\vec a\times \vec ba×b:

a⃗×b⃗=∣i^j^k^2−31325∣\vec a\times \vec b= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & -3 & 1\\ 3 & 2 & 5 \end{vmatrix}a×b=​i^23​j^​−32​k^15​​ =i^((−3)(5)−1⋅2)−j^(2⋅5−1⋅3)+k^(2⋅2−(−3)⋅3)=\hat i((-3)(5)-1\cdot 2)-\hat j(2\cdot 5-1\cdot 3)+\hat k(2\cdot 2-(-3)\cdot 3)=i^((−3)(5)−1⋅2)−j^​(2⋅5−1⋅3)+k^(2⋅2−(−3)⋅3) =−17i^−7j^+13k^.=-17\hat i-7\hat j+13\hat k.=−17i^−7j^​+13k^.

Given

a⃗×b⃗+d⃗=−18i^−3j^+12k^,\vec a\times \vec b+\vec d=-18\hat i-3\hat j+12\hat k,a×b+d=−18i^−3j^​+12k^,

so

d⃗=(−18,−3,12)−(−17,−7,13)=(−1,4,−1).\vec d=(-18,-3,12)-(-17,-7,13)=(-1,4,-1).d=(−18,−3,12)−(−17,−7,13)=(−1,4,−1).

Thus,

d⃗=−i^+4j^−k^.\vec d=-\hat i+4\hat j-\hat k.d=−i^+4j^​−k^.
  1. Now compute a⃗⋅d⃗\vec a\cdot \vec da⋅d
a⃗⋅d⃗=(2,−3,1)⋅(−1,4,−1)\vec a\cdot \vec d=(2,-3,1)\cdot(-1,4,-1)a⋅d=(2,−3,1)⋅(−1,4,−1) =2(−1)+(−3)(4)+1(−1)=−2−12−1=−15.=2(-1)+(-3)(4)+1(-1)=-2-12-1=-15.=2(−1)+(−3)(4)+1(−1)=−2−12−1=−15.

Therefore,

∣a⃗⋅d⃗∣=15.|\vec a\cdot \vec d|=15.∣a⋅d∣=15.
  1. Check with options

The value is

151515

So the correct option is A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They match.

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