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Vector Algebra question

2021 · 17 Mar · Shift 2 · Q37
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  5. /2021 · 17 Mar · Shift 2 · Q37

Vector Algebra question

2021 · 17 Mar · Shift 2 · Q37

JEE MainMathematicsVector AlgebraNumerical+4 / −1
Let x→\overrightarrow xx be a vector in the plane containing vectors a→=2i^−j^+k^\overrightarrow a = 2\widehat i - \widehat j + \widehat ka=2i−j​+k and b→=i^+2j^−k^\overrightarrow b = \widehat i + 2\widehat j - \widehat kb=i+2j​−k. If the vector x→\overrightarrow xx is perpendicular to (3i^+2j^−k^)\left( {3\widehat i + 2\widehat j - \widehat k} \right)(3i+2j​−k) and its projection on a→\overrightarrow aa is 1762{{17\sqrt 6 } \over 2}2176​​, then the value of ∣x→∣2|\overrightarrow x {|^2}∣x∣2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 486

  1. Since x⃗\vec xx lies in the plane containing a⃗\vec aa and b⃗\vec bb, write
x⃗=αa⃗+βb⃗\vec x=\alpha \vec a+\beta \vec bx=αa+βb

where

a⃗=(2,−1,1),b⃗=(1,2,−1).\vec a=(2,-1,1),\qquad \vec b=(1,2,-1).a=(2,−1,1),b=(1,2,−1).

So,

x⃗=α(2,−1,1)+β(1,2,−1)=(2α+β,−α+2β,α−β).\vec x=\alpha(2,-1,1)+\beta(1,2,-1)=(2\alpha+\beta, -\alpha+2\beta, \alpha-\beta).x=α(2,−1,1)+β(1,2,−1)=(2α+β,−α+2β,α−β).
  1. Given that x⃗\vec xx is perpendicular to (3,2,−1)(3,2,-1)(3,2,−1), so
x⃗⋅(3,2,−1)=0.\vec x\cdot(3,2,-1)=0.x⋅(3,2,−1)=0.

Substitute x⃗\vec xx:

(2α+β)3+(−α+2β)2+(α−β)(−1)=0.(2\alpha+\beta)3+(-\alpha+2\beta)2+( \alpha-\beta)(-1)=0.(2α+β)3+(−α+2β)2+(α−β)(−1)=0.

Simplify:

6α+3β−2α+4β−α+β=06\alpha+3\beta-2\alpha+4\beta-\alpha+\beta=06α+3β−2α+4β−α+β=0 3α+8β=0.3\alpha+8\beta=0.3α+8β=0.

Hence,

α=−83β.\alpha=-\frac{8}{3}\beta.α=−38​β.
  1. The projection of x⃗\vec xx on a⃗\vec aa means the scalar projection:
x⃗⋅a⃗∣a⃗∣=1762.\frac{\vec x\cdot \vec a}{|\vec a|}=\frac{17\sqrt6}{2}.∣a∣x⋅a​=2176​​.

Now,

∣a⃗∣=22+(−1)2+12=6.|\vec a|=\sqrt{2^2+(-1)^2+1^2}=\sqrt6.∣a∣=22+(−1)2+12​=6​.

Therefore,

x⃗⋅a⃗=1762⋅6=51.\vec x\cdot \vec a=\frac{17\sqrt6}{2}\cdot \sqrt6=51.x⋅a=2176​​⋅6​=51.
  1. Compute x⃗⋅a⃗\vec x\cdot \vec ax⋅a using x⃗=αa⃗+βb⃗\vec x=\alpha\vec a+\beta\vec bx=αa+βb:
x⃗⋅a⃗=α(a⃗⋅a⃗)+β(b⃗⋅a⃗).\vec x\cdot \vec a=\alpha(\vec a\cdot \vec a)+\beta(\vec b\cdot \vec a).x⋅a=α(a⋅a)+β(b⋅a).

Now,

a⃗⋅a⃗=6,\vec a\cdot \vec a=6,a⋅a=6, a⃗⋅b⃗=2(1)+(−1)(2)+(1)(−1)=2−2−1=−1.\vec a\cdot \vec b=2(1)+(-1)(2)+(1)(-1)=2-2-1=-1.a⋅b=2(1)+(−1)(2)+(1)(−1)=2−2−1=−1.

So,

6α−β=51.6\alpha-\beta=51.6α−β=51.

Using α=−83β\alpha=-\frac83\betaα=−38​β,

6(−83β)−β=516\left(-\frac83\beta\right)-\beta=516(−38​β)−β=51 −16β−β=51-16\beta-\beta=51−16β−β=51 −17β=51-17\beta=51−17β=51 β=−3.\beta=-3.β=−3.

Thus,

α=8.\alpha=8.α=8.
  1. Hence,
x⃗=8a⃗−3b⃗.\vec x=8\vec a-3\vec b.x=8a−3b.

Compute:

8a⃗=(16,−8,8),3b⃗=(3,6,−3),8\vec a=(16,-8,8),\qquad 3\vec b=(3,6,-3),8a=(16,−8,8),3b=(3,6,−3),

so

x⃗=(16,−8,8)−(3,6,−3)=(13,−14,11).\vec x=(16,-8,8)-(3,6,-3)=(13,-14,11).x=(16,−8,8)−(3,6,−3)=(13,−14,11).
  1. Now find ∣x⃗∣2|\vec x|^2∣x∣2:
∣x⃗∣2=132+(−14)2+112=169+196+121=486.|\vec x|^2=13^2+(-14)^2+11^2=169+196+121=486.∣x∣2=132+(−14)2+112=169+196+121=486.

Therefore,

486\boxed{486}486​
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