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Vector Algebra question

2023 · 1 Feb · Shift 2 · Q36
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  5. /2023 · 1 Feb · Shift 2 · Q36

Vector Algebra question

2023 · 1 Feb · Shift 2 · Q36

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a⃗=2i^−7j^+5k^,b⃗=i^+k^\vec{a}=2 \hat{i}-7 \hat{j}+5 \hat{k}, \vec{b}=\hat{i}+\hat{k}a=2i^−7j^​+5k^,b=i^+k^ and c⃗=i^+2j^−3k^\vec{c}=\hat{i}+2 \hat{j}-3 \hat{k}c=i^+2j^​−3k^ be three given vectors. If r→\overrightarrow{\mathrm{r}}r is a vector such that r⃗×a⃗=c⃗×a⃗\vec{r} \times \vec{a}=\vec{c} \times \vec{a}r×a=c×a and r⃗⋅b⃗=0\vec{r} \cdot \vec{b}=0r⋅b=0, then ∣r⃗∣|\vec{r}|∣r∣ is equal to :
  1. A
    117\frac{11}{7}711​
  2. B
    1152\frac{11}{5} \sqrt{2}511​2​
  3. C
    9147\frac{\sqrt{914}}{7}7914​​
  4. D
    1172\frac{11}{7} \sqrt{2}711​2​
View written solutionFree

Correct answer: D

  1. Given vectors
a⃗=(2,−7,5),b⃗=(1,0,1),c⃗=(1,2,−3)\vec a=(2,-7,5),\quad \vec b=(1,0,1),\quad \vec c=(1,2,-3)a=(2,−7,5),b=(1,0,1),c=(1,2,−3)

We need a vector r⃗\vec rr such that:

r⃗×a⃗=c⃗×a⃗\vec r\times \vec a=\vec c\times \vec ar×a=c×a

and

r⃗⋅b⃗=0\vec r\cdot \vec b=0r⋅b=0

We must find ∣r⃗∣|\vec r|∣r∣.

  1. Use the cross product condition

From

r⃗×a⃗=c⃗×a⃗\vec r\times \vec a=\vec c\times \vec ar×a=c×a

we get

(r⃗−c⃗)×a⃗=0⃗(\vec r-\vec c)\times \vec a=\vec 0(r−c)×a=0

This means r⃗−c⃗\vec r-\vec cr−c is parallel to a⃗\vec aa. Hence,

r⃗=c⃗+λa⃗\vec r=\vec c+\lambda \vec ar=c+λa

for some scalar λ\lambdaλ.

So,

r⃗=(1,2,−3)+λ(2,−7,5)\vec r=(1,2,-3)+\lambda(2,-7,5)r=(1,2,−3)+λ(2,−7,5)

Thus,

r⃗=(1+2λ, 2−7λ, −3+5λ)\vec r=(1+2\lambda,\ 2-7\lambda,\ -3+5\lambda)r=(1+2λ, 2−7λ, −3+5λ)
  1. Use the dot product condition

Given

r⃗⋅b⃗=0,\vec r\cdot \vec b=0,r⋅b=0,

with b⃗=(1,0,1)\vec b=(1,0,1)b=(1,0,1).

So,

(1+2λ, 2−7λ, −3+5λ)⋅(1,0,1)=0(1+2\lambda,\ 2-7\lambda,\ -3+5\lambda)\cdot (1,0,1)=0(1+2λ, 2−7λ, −3+5λ)⋅(1,0,1)=0 (1+2λ)+(−3+5λ)=0(1+2\lambda)+(-3+5\lambda)=0(1+2λ)+(−3+5λ)=0 −2+7λ=0-2+7\lambda=0−2+7λ=0 λ=27\lambda=\frac{2}{7}λ=72​

Therefore,

r⃗=c⃗+27a⃗\vec r=\vec c+\frac{2}{7}\vec ar=c+72​a

Compute components:

r⃗=(1+2⋅27, 2−7⋅27, −3+5⋅27)\vec r=\left(1+2\cdot\frac{2}{7},\ 2-7\cdot\frac{2}{7},\ -3+5\cdot\frac{2}{7}\right)r=(1+2⋅72​, 2−7⋅72​, −3+5⋅72​) r⃗=(1+47, 2−2, −3+107)\vec r=\left(1+\frac{4}{7},\ 2-2,\ -3+\frac{10}{7}\right)r=(1+74​, 2−2, −3+710​) r⃗=(117, 0, −117)\vec r=\left(\frac{11}{7},\ 0,\ -\frac{11}{7}\right)r=(711​, 0, −711​)
  1. Find magnitude of r⃗\vec rr
∣r⃗∣=(117)2+02+(−117)2|\vec r|=\sqrt{\left(\frac{11}{7}\right)^2+0^2+\left(-\frac{11}{7}\right)^2}∣r∣=(711​)2+02+(−711​)2​ ∣r⃗∣=2(117)2|\vec r|=\sqrt{2\left(\frac{11}{7}\right)^2}∣r∣=2(711​)2​ ∣r⃗∣=1172|\vec r|=\frac{11}{7}\sqrt{2}∣r∣=711​2​
  1. Match with options
∣r⃗∣=1172|\vec r|=\frac{11}{7}\sqrt{2}∣r∣=711​2​

So the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.

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