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Vector Algebra question

2022 · 27 Jul · Shift 1 · Q34
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  5. /2022 · 27 Jul · Shift 1 · Q34

Vector Algebra question

2022 · 27 Jul · Shift 1 · Q34

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a⃗=αi^+j^+βk^\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}a=αi^+j^​+βk^ and b⃗=3i^−5j^+4k^\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}b=3i^−5j^​+4k^ be two vectors, such that a⃗×b⃗=−i^+9j^+12k^\vec{a} \times \vec{b}=-\hat{i}+9 \hat{j}+12 \hat{k}a×b=−i^+9j^​+12k^. Then the projection of b⃗−2a⃗\vec{b}-2 \vec{a}b−2a on b⃗+a⃗\vec{b}+\vec{a}b+a is equal to :
  1. A
    2
  2. B
    395\frac{39}{5}539​
  3. C
    9
  4. D
    465\frac{46}{5}546​
View written solutionFree

Correct answer: D

  1. Write the vectors
a⃗=(α,1,β),b⃗=(3,−5,4)\vec a=(\alpha,1,\beta),\qquad \vec b=(3,-5,4)a=(α,1,β),b=(3,−5,4)

Given:

a⃗×b⃗=−i^+9j^+12k^=(−1,9,12)\vec a\times \vec b=-\hat i+9\hat j+12\hat k=(-1,9,12)a×b=−i^+9j^​+12k^=(−1,9,12)
  1. Compute a⃗×b⃗\vec a\times \vec ba×b in terms of α,β\alpha,\betaα,β

Using the determinant,

a⃗×b⃗=∣i^j^k^α1β3−54∣\vec a\times \vec b= \begin{vmatrix} \hat i & \hat j & \hat k\\ \alpha & 1 & \beta\\ 3 & -5 & 4 \end{vmatrix}a×b=​i^α3​j^​1−5​k^β4​​

So,

a⃗×b⃗=i^(1⋅4−β(−5))−j^(α⋅4−β⋅3)+k^(α(−5)−1⋅3)\vec a\times \vec b= \hat i(1\cdot 4-\beta(-5)) -\hat j(\alpha\cdot 4-\beta\cdot 3) +\hat k(\alpha(-5)-1\cdot 3)a×b=i^(1⋅4−β(−5))−j^​(α⋅4−β⋅3)+k^(α(−5)−1⋅3) =(4+5β)i^−(4α−3β)j^+(−5α−3)k^= (4+5\beta)\hat i-(4\alpha-3\beta)\hat j+(-5\alpha-3)\hat k=(4+5β)i^−(4α−3β)j^​+(−5α−3)k^

This equals

−i^+9j^+12k^-\hat i+9\hat j+12\hat k−i^+9j^​+12k^

Hence, by comparing components:

4+5β=−14+5\beta=-14+5β=−1 −(4α−3β)=9-(4\alpha-3\beta)=9−(4α−3β)=9 −5α−3=12-5\alpha-3=12−5α−3=12
  1. Solve for α\alphaα and β\betaβ

From the first equation:

5β=−5  ⟹  β=−15\beta=-5\implies \beta=-15β=−5⟹β=−1

From the third equation:

−5α=15  ⟹  α=−3-5\alpha=15\implies \alpha=-3−5α=15⟹α=−3

Check in the second equation:

−(4(−3)−3(−1))=−(−12+3)=9-(4(-3)-3(-1))= -(-12+3)=9−(4(−3)−3(−1))=−(−12+3)=9

Correct.

Thus,

a⃗=−3i^+j^−k^\vec a=-3\hat i+\hat j-\hat ka=−3i^+j^​−k^
  1. Find b⃗−2a⃗\vec b-2\vec ab−2a and b⃗+a⃗\vec b+\vec ab+a

First,

2a⃗=2(−3,1,−1)=(−6,2,−2)2\vec a=2(-3,1,-1)=(-6,2,-2)2a=2(−3,1,−1)=(−6,2,−2)

So,

b⃗−2a⃗=(3,−5,4)−(−6,2,−2)=(9,−7,6)\vec b-2\vec a=(3,-5,4)-(-6,2,-2)=(9,-7,6)b−2a=(3,−5,4)−(−6,2,−2)=(9,−7,6)

Also,

b⃗+a⃗=(3,−5,4)+(−3,1,−1)=(0,−4,3)\vec b+\vec a=(3,-5,4)+(-3,1,-1)=(0,-4,3)b+a=(3,−5,4)+(−3,1,−1)=(0,−4,3)
  1. Projection of b⃗−2a⃗\vec b-2\vec ab−2a on b⃗+a⃗\vec b+\vec ab+a

The scalar projection of vector u⃗\vec uu on vector v⃗\vec vv is

u⃗⋅v⃗∣v⃗∣\frac{\vec u\cdot \vec v}{|\vec v|}∣v∣u⋅v​

Here,

u⃗=b⃗−2a⃗=(9,−7,6),v⃗=b⃗+a⃗=(0,−4,3)\vec u=\vec b-2\vec a=(9,-7,6),\qquad \vec v=\vec b+\vec a=(0,-4,3)u=b−2a=(9,−7,6),v=b+a=(0,−4,3)

Compute dot product:

u⃗⋅v⃗=9⋅0+(−7)(−4)+6⋅3=28+18=46\vec u\cdot \vec v=9\cdot 0+(-7)(-4)+6\cdot 3=28+18=46u⋅v=9⋅0+(−7)(−4)+6⋅3=28+18=46

Magnitude of v⃗\vec vv:

∣v⃗∣=02+(−4)2+32=16+9=5|\vec v|=\sqrt{0^2+(-4)^2+3^2}=\sqrt{16+9}=5∣v∣=02+(−4)2+32​=16+9​=5

Therefore projection is

465\frac{46}{5}546​
  1. Match with options
465\frac{46}{5}546​

So the correct option is D.

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