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Vector Algebra question

2024 · 4 Apr · Shift 2 · Q50
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  5. /2024 · 4 Apr · Shift 2 · Q50

Vector Algebra question

2024 · 4 Apr · Shift 2 · Q50

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a⃗=i^+j^+k^,b⃗=2i^+4j^−5k^\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}a=i^+j^​+k^,b=2i^+4j^​−5k^ and c⃗=xi^+2j^+3k^,x∈R\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in \mathbb{R}c=xi^+2j^​+3k^,x∈R. If d⃗\vec{d}d is the unit vector in the direction of b⃗+c⃗\vec{b}+\vec{c}b+c such that a⃗⋅d⃗=1\vec{a} \cdot \vec{d}=1a⋅d=1, then (a⃗×b⃗)⋅c⃗(\vec{a} \times \vec{b}) \cdot \vec{c}(a×b)⋅c is equal to
  1. A
    3
  2. B
    9
  3. C
    11
  4. D
    6
View written solutionFree

Correct answer: C

  1. Given vectors
a⃗=(1,1,1),b⃗=(2,4,−5),c⃗=(x,2,3)\vec a=(1,1,1),\quad \vec b=(2,4,-5),\quad \vec c=(x,2,3)a=(1,1,1),b=(2,4,−5),c=(x,2,3)

We are told that d⃗\vec dd is the unit vector in the direction of b⃗+c⃗\vec b+\vec cb+c and satisfies

a⃗⋅d⃗=1.\vec a\cdot \vec d=1.a⋅d=1.
  1. Find b⃗+c⃗\vec b+\vec cb+c
b⃗+c⃗=(2+x,6,−2).\vec b+\vec c=(2+x,6,-2).b+c=(2+x,6,−2).

Since d⃗\vec dd is the unit vector in this direction,

d⃗=b⃗+c⃗∣b⃗+c⃗∣=(x+2,6,−2)(x+2)2+62+(−2)2.\vec d=\frac{\vec b+\vec c}{|\vec b+\vec c|} =\frac{(x+2,6,-2)}{\sqrt{(x+2)^2+6^2+(-2)^2}}.d=∣b+c∣b+c​=(x+2)2+62+(−2)2​(x+2,6,−2)​.

So,

a⃗⋅d⃗=(1,1,1)⋅(x+2,6,−2)(x+2)2+36+4.\vec a\cdot \vec d =\frac{(1,1,1)\cdot (x+2,6,-2)}{\sqrt{(x+2)^2+36+4}}.a⋅d=(x+2)2+36+4​(1,1,1)⋅(x+2,6,−2)​.

The numerator is

(x+2)+6−2=x+6.(x+2)+6-2=x+6.(x+2)+6−2=x+6.

Thus,

a⃗⋅d⃗=x+6(x+2)2+40=1.\vec a\cdot \vec d=\frac{x+6}{\sqrt{(x+2)^2+40}}=1.a⋅d=(x+2)2+40​x+6​=1.

Therefore,

x+6=(x+2)2+40.x+6=\sqrt{(x+2)^2+40}.x+6=(x+2)2+40​.

Squaring both sides,

(x+6)2=(x+2)2+40.(x+6)^2=(x+2)^2+40.(x+6)2=(x+2)2+40.

Expand:

x2+12x+36=x2+4x+4+40=x2+4x+44.x^2+12x+36=x^2+4x+4+40=x^2+4x+44.x2+12x+36=x2+4x+4+40=x2+4x+44.

So,

12x+36=4x+4412x+36=4x+4412x+36=4x+44 8x=88x=88x=8 x=1.x=1.x=1.

Hence,

c⃗=(1,2,3).\vec c=(1,2,3).c=(1,2,3).
  1. Compute a⃗×b⃗\vec a\times \vec ba×b
a⃗×b⃗=∣i^j^k^11124−5∣\vec a\times \vec b= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1&1&1\\ 2&4&-5 \end{vmatrix}a×b=​i^12​j^​14​k^1−5​​ =i^∣114−5∣−j^∣112−5∣+k^∣1124∣=\hat i\begin{vmatrix}1&1\\4&-5\end{vmatrix} -\hat j\begin{vmatrix}1&1\\2&-5\end{vmatrix} +\hat k\begin{vmatrix}1&1\\2&4\end{vmatrix}=i^​14​1−5​​−j^​​12​1−5​​+k^​12​14​​ =i^(−5−4)−j^(−5−2)+k^(4−2)=\hat i(-5-4)-\hat j(-5-2)+\hat k(4-2)=i^(−5−4)−j^​(−5−2)+k^(4−2) =−9i^+7j^+2k^.=-9\hat i+7\hat j+2\hat k.=−9i^+7j^​+2k^.

So,

a⃗×b⃗=(−9,7,2).\vec a\times \vec b=(-9,7,2).a×b=(−9,7,2).
  1. Compute (a⃗×b⃗)⋅c⃗(\vec a\times \vec b)\cdot \vec c(a×b)⋅c

With c⃗=(1,2,3)\vec c=(1,2,3)c=(1,2,3),

(a⃗×b⃗)⋅c⃗=(−9,7,2)⋅(1,2,3).(\vec a\times \vec b)\cdot \vec c=(-9,7,2)\cdot(1,2,3).(a×b)⋅c=(−9,7,2)⋅(1,2,3). =−9+14+6=11.=-9+14+6=11.=−9+14+6=11.
  1. Match with options
(a⃗×b⃗)⋅c⃗=11(\vec a\times \vec b)\cdot \vec c=11(a×b)⋅c=11

So the correct option is C.

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.

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