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Vector Algebra question

2024 · 31 Jan · Shift 1 · Q42
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  5. /2024 · 31 Jan · Shift 1 · Q42

Vector Algebra question

2024 · 31 Jan · Shift 1 · Q42

JEE MainMathematicsVector AlgebraMCQ+4 / −1
The distance of the point Q(0,2,−2)Q(0,2,-2)Q(0,2,−2) form the line passing through the point P(5,−4,3)P(5,-4, 3)P(5,−4,3) and perpendicular to the lines r⃗=(−3i^+2k^)+λ(2i^+3j^+5k^),λ∈R\vec{r}=(-3 \hat{i}+2 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+5 \hat{k}), \lambda \in \mathbb{R}r=(−3i^+2k^)+λ(2i^+3j^​+5k^),λ∈R and r⃗=(i^−2j^+k^)+μ(−i^+3j^+2k^),μ∈R\vec{r}=(\hat{i}-2 \hat{j}+\hat{k})+\mu(-\hat{i}+3 \hat{j}+2 \hat{k}), \mu \in \mathbb{R}r=(i^−2j^​+k^)+μ(−i^+3j^​+2k^),μ∈R is :
  1. A
    74\sqrt{74}74​
  2. B
    86\sqrt{86}86​
  3. C
    54\sqrt{54}54​
  4. D
    20\sqrt{20}20​
View written solutionFree

Correct answer: A

  1. Find the direction vector of the required line

The required line passes through P(5,−4,3)P(5,-4,3)P(5,−4,3) and is perpendicular to both given lines. So its direction vector must be perpendicular to the direction vectors of both lines.

The direction vectors are: a⃗=(2,3,5),b⃗=(−1,3,2)\vec{a}=(2,3,5), \qquad \vec{b}=(-1,3,2)a=(2,3,5),b=(−1,3,2)

Hence the required direction vector is d⃗=a⃗×b⃗\vec{d}=\vec{a}\times \vec{b}d=a×b

Compute the cross product:

d⃗=∣i^j^k^235−132∣\vec{d}=\begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{vmatrix}d=​i^2−1​j^​33​k^52​​ d⃗=i^(3⋅2−5⋅3)−j^(2⋅2−5⋅(−1))+k^(2⋅3−3⋅(−1))\vec{d}=\hat i(3\cdot 2-5\cdot 3)-\hat j(2\cdot 2-5\cdot(-1))+\hat k(2\cdot 3-3\cdot(-1))d=i^(3⋅2−5⋅3)−j^​(2⋅2−5⋅(−1))+k^(2⋅3−3⋅(−1)) d⃗=i^(6−15)−j^(4+5)+k^(6+3)\vec{d}=\hat i(6-15)-\hat j(4+5)+\hat k(6+3)d=i^(6−15)−j^​(4+5)+k^(6+3) d⃗=(−9,−9,9)\vec{d}=(-9,-9,9)d=(−9,−9,9)

So a simpler direction vector is d⃗=(1,1,−1)\vec{d}=(1,1,-1)d=(1,1,−1)

Thus the required line is r⃗=(5,−4,3)+t(1,1,−1)\vec r=(5,-4,3)+t(1,1,-1)r=(5,−4,3)+t(1,1,−1)

  1. Use point-to-line distance formula

We need the distance of point Q(0,2,−2)Q(0,2,-2)Q(0,2,−2) from this line.

Take P=(5,−4,3),Q=(0,2,−2)P=(5,-4,3), \qquad Q=(0,2,-2)P=(5,−4,3),Q=(0,2,−2)

Then PQ→=Q−P=(0−5,2−(−4),−2−3)=(−5,6,−5)\overrightarrow{PQ}=Q-P=(0-5,2-(-4),-2-3)=(-5,6,-5)PQ​=Q−P=(0−5,2−(−4),−2−3)=(−5,6,−5)

Distance from point to line is

Distance=∣PQ→×d⃗∣∣d⃗∣\text{Distance}=\frac{|\overrightarrow{PQ}\times \vec d|}{|\vec d|}Distance=∣d∣∣PQ​×d∣​

with d⃗=(1,1,−1)\vec d=(1,1,-1)d=(1,1,−1).

  1. Compute the cross product
PQ→×d⃗=∣i^j^k^−56−511−1∣\overrightarrow{PQ}\times \vec d= \begin{vmatrix} \hat i & \hat j & \hat k \\ -5 & 6 & -5 \\ 1 & 1 & -1 \end{vmatrix}PQ​×d=​i^−51​j^​61​k^−5−1​​ =i^(6⋅(−1)−(−5)⋅1)−j^((−5)⋅(−1)−(−5)⋅1)+k^((−5)⋅1−6⋅1)=\hat i(6\cdot(-1)-(-5)\cdot 1)-\hat j((-5)\cdot(-1)-(-5)\cdot 1)+\hat k((-5)\cdot 1-6\cdot 1)=i^(6⋅(−1)−(−5)⋅1)−j^​((−5)⋅(−1)−(−5)⋅1)+k^((−5)⋅1−6⋅1) =i^(−6+5)−j^(5+5)+k^(−5−6)=\hat i(-6+5)-\hat j(5+5)+\hat k(-5-6)=i^(−6+5)−j^​(5+5)+k^(−5−6) =(−1,−10,−11)=(-1,-10,-11)=(−1,−10,−11)

So

∣PQ→×d⃗∣=(−1)2+(−10)2+(−11)2=1+100+121=222|\overrightarrow{PQ}\times \vec d|=\sqrt{(-1)^2+(-10)^2+(-11)^2}= \sqrt{1+100+121}= \sqrt{222}∣PQ​×d∣=(−1)2+(−10)2+(−11)2​=1+100+121​=222​

Also,

∣d⃗∣=12+12+(−1)2=3|\vec d|=\sqrt{1^2+1^2+(-1)^2}=\sqrt{3}∣d∣=12+12+(−1)2​=3​

Therefore,

Distance=2223=74\text{Distance}=\frac{\sqrt{222}}{\sqrt{3}}=\sqrt{74}Distance=3​222​​=74​
  1. Compare with options

74\sqrt{74}74​ matches Option A.

  1. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.

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