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Vector Algebra question

2023 · 31 Jan · Shift 2 · Q23
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  5. /2023 · 31 Jan · Shift 2 · Q23

Vector Algebra question

2023 · 31 Jan · Shift 2 · Q23

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a⃗=i^+2j^+3k^,b⃗=i^−j^+2k^\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k}a=i^+2j^​+3k^,b=i^−j^​+2k^ and c⃗=5i^−3j^+3k^\vec{c}=5 \hat{i}-3 \hat{j}+3 \hat{k}c=5i^−3j^​+3k^ be three vectors. If r⃗\vec{r}r is a vector such that, r⃗×b⃗=c⃗×b⃗\vec{r} \times \vec{b}=\vec{c} \times \vec{b}r×b=c×b and r⃗⋅a⃗=0\vec{r} \cdot \vec{a}=0r⋅a=0, then 25∣r⃗∣225|\vec{r}|^{2}25∣r∣2 is equal to :
  1. A
    336
  2. B
    449
  3. C
    339
  4. D
    560
View written solutionFree

Correct answer: C

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

We need r⃗\vec rr such that

r⃗×b⃗=c⃗×b⃗\vec r\times \vec b=\vec c\times \vec br×b=c×b

and

r⃗⋅a⃗=0.\vec r\cdot \vec a=0.r⋅a=0.

We must find 25∣r⃗∣225|\vec r|^225∣r∣2.

  1. Use the cross product condition

From

r⃗×b⃗=c⃗×b⃗\vec r\times \vec b=\vec c\times \vec br×b=c×b

we get

(r⃗−c⃗)×b⃗=0⃗.(\vec r-\vec c)\times \vec b=\vec 0.(r−c)×b=0.

Hence r⃗−c⃗\vec r-\vec cr−c is parallel to b⃗\vec bb. So,

r⃗=c⃗+λb⃗\vec r=\vec c+\lambda \vec br=c+λb

for some scalar λ\lambdaλ.

Substitute the given vectors:

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

So,

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

Given

r⃗⋅a⃗=0,\vec r\cdot \vec a=0,r⋅a=0,

so

(5+λ,−3−λ,3+2λ)⋅(1,2,3)=0.(5+\lambda,-3-\lambda,3+2\lambda)\cdot(1,2,3)=0.(5+λ,−3−λ,3+2λ)⋅(1,2,3)=0.

Compute:

(5+λ)+2(−3−λ)+3(3+2λ)=0.(5+\lambda)+2(-3-\lambda)+3(3+2\lambda)=0.(5+λ)+2(−3−λ)+3(3+2λ)=0. 5+λ−6−2λ+9+6λ=0.5+\lambda-6-2\lambda+9+6\lambda=0.5+λ−6−2λ+9+6λ=0. 8+5λ=0.8+5\lambda=0.8+5λ=0.

Thus,

λ=−85.\lambda=-\frac{8}{5}.λ=−58​.

Therefore,

r⃗=c⃗−85b⃗.\vec r=\vec c-\frac{8}{5}\vec b.r=c−58​b.

Now compute components:

r⃗=(5−85, −3+85, 3−165)\vec r=\left(5-\frac{8}{5},\,-3+\frac{8}{5},\,3-\frac{16}{5}\right)r=(5−58​,−3+58​,3−516​) r⃗=(175, −75, −15).\vec r=\left(\frac{17}{5},\,-\frac{7}{5},\,-\frac{1}{5}\right).r=(517​,−57​,−51​).
  1. Find ∣r⃗∣2|\vec r|^2∣r∣2
∣r⃗∣2=(175)2+(−75)2+(−15)2|\vec r|^2=\left(\frac{17}{5}\right)^2+\left(-\frac{7}{5}\right)^2+\left(-\frac{1}{5}\right)^2∣r∣2=(517​)2+(−57​)2+(−51​)2 =28925+4925+125=\frac{289}{25}+\frac{49}{25}+\frac{1}{25}=25289​+2549​+251​ =33925.=\frac{339}{25}.=25339​.

Hence,

25∣r⃗∣2=25⋅33925=339.25|\vec r|^2=25\cdot \frac{339}{25}=339.25∣r∣2=25⋅25339​=339.
  1. Check options

The value is

339\boxed{339}339​

So the correct option is C.

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