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Vector Algebra question

2022 · 25 Jun · Shift 2 · Q42
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Vector Algebra question

2022 · 25 Jun · Shift 2 · Q42

JEE MainMathematicsVector AlgebraNumerical+4 / −1
Let b→=i^+j^+λk^\overrightarrow b = \widehat i + \widehat j + \lambda \widehat kb=i+j​+λk, λ∈\lambda\inλ∈ R. If a→\overrightarrow aa is a vector such that a→×b→=13i^−j^−4k^\overrightarrow a \times \overrightarrow b = 13\widehat i - \widehat j - 4\widehat ka×b=13i−j​−4k and a→ . b→+21=0\overrightarrow a \,.\,\overrightarrow b + 21 = 0a.b+21=0, then (b→−a→). (k^−j^)+(b→+a→). (i^−k^)\left( {\overrightarrow b - \overrightarrow a } \right).\,\left( {\widehat k - \widehat j} \right) + \left( {\overrightarrow b + \overrightarrow a } \right).\,\left( {\widehat i - \widehat k} \right)(b−a).(k−j​)+(b+a).(i−k) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 14

Let b⃗=i^+j^+λk^=(1,1,λ)\vec b=\hat i+\hat j+\lambda \hat k=(1,1,\lambda)b=i^+j^​+λk^=(1,1,λ) and let a⃗=(x,y,z).\vec a=(x,y,z).a=(x,y,z).

We are given:

  1. a⃗×b⃗=13i^−j^−4k^=(13,−1,−4)\vec a\times \vec b=13\hat i-\hat j-4\hat k=(13,-1,-4)a×b=13i^−j^​−4k^=(13,−1,−4)
  2. a⃗⋅b⃗+21=0  ⟹  a⃗⋅b⃗=−21\vec a\cdot \vec b+21=0\implies \vec a\cdot \vec b=-21a⋅b+21=0⟹a⋅b=−21

We must find (b⃗−a⃗)⋅(k^−j^)+(b⃗+a⃗)⋅(i^−k^).(\vec b-\vec a)\cdot(\hat k-\hat j)+(\vec b+\vec a)\cdot(\hat i-\hat k).(b−a)⋅(k^−j^​)+(b+a)⋅(i^−k^).

1. Use the cross product condition

Compute

a⃗×b⃗=∣i^j^k^xyz11λ∣\vec a\times \vec b= \begin{vmatrix} \hat i & \hat j & \hat k\\ x & y & z\\ 1 & 1 & \lambda \end{vmatrix}a×b=​i^x1​j^​y1​k^zλ​​

So,

a⃗×b⃗=i^(yλ−z)−j^(xλ−z)+k^(x−y).\vec a\times \vec b= \hat i(y\lambda-z)-\hat j(x\lambda-z)+\hat k(x-y).a×b=i^(yλ−z)−j^​(xλ−z)+k^(x−y).

Comparing with (13,−1,−4)(13,-1,-4)(13,−1,−4), we get:

yλ−z=13...(1)y\lambda-z=13 \quad ...(1)yλ−z=13...(1) −(xλ−z)=−1  ⟹  xλ−z=1...(2)-(x\lambda-z)=-1 \implies x\lambda-z=1 \quad ...(2)−(xλ−z)=−1⟹xλ−z=1...(2) x−y=−4...(3)x-y=-4 \quad ...(3)x−y=−4...(3)

From (3), x=y−4.x=y-4. x=y−4.

Subtract (2) from (1): yλ−xλ=12y\lambda-x\lambda=12yλ−xλ=12 λ(y−x)=12.\lambda(y-x)=12.λ(y−x)=12.

But from (3), x−y=−4  ⟹  y−x=4.x-y=-4 \implies y-x=4.x−y=−4⟹y−x=4. Hence, 4λ=12  ⟹  λ=3.4\lambda=12 \implies \lambda=3.4λ=12⟹λ=3.

So, b⃗=(1,1,3).\vec b=(1,1,3).b=(1,1,3).

2. Find components of a⃗\vec aa

Using (2): xλ−z=1x\lambda-z=1xλ−z=1 3x−z=1  ⟹  z=3x−1.3x-z=1 \implies z=3x-1. 3x−z=1⟹z=3x−1.

Using (3): y=x+4.y=x+4. y=x+4.

Now use the dot product condition: a⃗⋅b⃗=−21\vec a\cdot \vec b=-21a⋅b=−21 x+y+3z=−21.x+y+3z=-21.x+y+3z=−21.

Substitute y=x+4y=x+4y=x+4 and z=3x−1z=3x-1z=3x−1: x+(x+4)+3(3x−1)=−21x+(x+4)+3(3x-1)=-21x+(x+4)+3(3x−1)=−21 x+x+4+9x−3=−21x+x+4+9x-3=-21x+x+4+9x−3=−21 11x+1=−2111x+1=-2111x+1=−21 11x=−2211x=-2211x=−22 x=−2.x=-2.x=−2.

Then y=x+4=2,y=x+4=2,y=x+4=2, z=3x−1=−6−1=−7.z=3x-1=-6-1=-7.z=3x−1=−6−1=−7.

Thus, a⃗=(−2,2,−7).\vec a=(-2,2,-7).a=(−2,2,−7).

3. Evaluate the required expression

We need E=(b⃗−a⃗)⋅(k^−j^)+(b⃗+a⃗)⋅(i^−k^).E=(\vec b-\vec a)\cdot(\hat k-\hat j)+(\vec b+\vec a)\cdot(\hat i-\hat k).E=(b−a)⋅(k^−j^​)+(b+a)⋅(i^−k^).

First, k^−j^=(0,−1,1),i^−k^=(1,0,−1).\hat k-\hat j=(0,-1,1), \qquad \hat i-\hat k=(1,0,-1).k^−j^​=(0,−1,1),i^−k^=(1,0,−1).

Also, b⃗−a⃗=(1−(−2),1−2,3−(−7))=(3,−1,10),\vec b-\vec a=(1-(-2),1-2,3-(-7))=(3,-1,10),b−a=(1−(−2),1−2,3−(−7))=(3,−1,10), b⃗+a⃗=(1+(−2),1+2,3+(−7))=(−1,3,−4).\vec b+\vec a=(1+(-2),1+2,3+(-7))=(-1,3,-4).b+a=(1+(−2),1+2,3+(−7))=(−1,3,−4).

Now,

(b⃗−a⃗)⋅(k^−j^)=(3,−1,10)⋅(0,−1,1)=0+1+10=11.(\vec b-\vec a)\cdot(\hat k-\hat j) =(3,-1,10)\cdot(0,-1,1)=0+1+10=11.(b−a)⋅(k^−j^​)=(3,−1,10)⋅(0,−1,1)=0+1+10=11.

And,

(b⃗+a⃗)⋅(i^−k^)=(−1,3,−4)⋅(1,0,−1)=−1+0+4=3.(\vec b+\vec a)\cdot(\hat i-\hat k) =(-1,3,-4)\cdot(1,0,-1)=-1+0+4=3.(b+a)⋅(i^−k^)=(−1,3,−4)⋅(1,0,−1)=−1+0+4=3.

Therefore, E=11+3=14.E=11+3=14.E=11+3=14.

4. Final answer

14\boxed{14}14​

This matches the stored correct answer.

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