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Vector Algebra question

2019 · 12 Apr · Shift 1 · Q37
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  5. /2019 · 12 Apr · Shift 1 · Q37

Vector Algebra question

2019 · 12 Apr · Shift 1 · Q37

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a→=3i^+2j^+2k^\overrightarrow a = 3\widehat i + 2\widehat j + 2\widehat ka=3i+2j​+2k and b→=i^+2j^−2k^\overrightarrow b = \widehat i + 2\widehat j - 2\widehat kb=i+2j​−2k be two vectors. If a vector perpendicular to both the vectors a→+b→\overrightarrow a + \overrightarrow ba+b and a→−b→\overrightarrow a - \overrightarrow ba−b has the magnitude 12 then one such vector is :
  1. A
    4(2i^−2j^−k^)4\left( {2\widehat i - 2\widehat j - \widehat k} \right)4(2i−2j​−k)
  2. B
    4(−2i^−2j^+k^)4\left( { - 2\widehat i - 2\widehat j + \widehat k} \right)4(−2i−2j​+k)
  3. C
    4(2i^+2j^+k^)4\left( {2\widehat i + 2\widehat j + \widehat k} \right)4(2i+2j​+k)
  4. D
    4(2i^+2j^−k^)4\left( {2\widehat i + 2\widehat j - \widehat k} \right)4(2i+2j​−k)
View written solutionFree

Correct answer: A

  1. We need a vector perpendicular to both a⃗+b⃗\vec a+\vec ba+b and a⃗−b⃗\vec a-\vec ba−b.

Given: a⃗=3i^+2j^+2k^,b⃗=i^+2j^−2k^\vec a=3\hat i+2\hat j+2\hat k,\qquad \vec b=\hat i+2\hat j-2\hat ka=3i^+2j^​+2k^,b=i^+2j^​−2k^

  1. Compute a⃗+b⃗\vec a+\vec ba+b and a⃗−b⃗\vec a-\vec ba−b.

a⃗+b⃗=(3+1)i^+(2+2)j^+(2−2)k^=4i^+4j^\vec a+\vec b=(3+1)\hat i+(2+2)\hat j+(2-2)\hat k=4\hat i+4\hat ja+b=(3+1)i^+(2+2)j^​+(2−2)k^=4i^+4j^​

So, a⃗+b⃗=4i^+4j^\vec a+\vec b=4\hat i+4\hat ja+b=4i^+4j^​

Next, a⃗−b⃗=(3−1)i^+(2−2)j^+(2−(−2))k^=2i^+4k^\vec a-\vec b=(3-1)\hat i+(2-2)\hat j+(2-(-2))\hat k=2\hat i+4\hat ka−b=(3−1)i^+(2−2)j^​+(2−(−2))k^=2i^+4k^

So, a⃗−b⃗=2i^+4k^\vec a-\vec b=2\hat i+4\hat ka−b=2i^+4k^

  1. A vector perpendicular to both is parallel to their cross product: (a⃗+b⃗)×(a⃗−b⃗)\left(\vec a+\vec b\right)\times\left(\vec a-\vec b\right)(a+b)×(a−b)

Now,

(4,4,0)×(2,0,4)=∣i^j^k^440204∣=i^(4⋅4−0⋅0)−j^(4⋅4−0⋅2)+k^(4⋅0−4⋅2)=16i^−16j^−8k^\begin{aligned} (4,4,0)\times(2,0,4) &=\begin{vmatrix} \hat i & \hat j & \hat k\\ 4 & 4 & 0\\ 2 & 0 & 4 \end{vmatrix}\\ &=\hat i(4\cdot 4-0\cdot 0)-\hat j(4\cdot 4-0\cdot 2)+\hat k(4\cdot 0-4\cdot 2)\\ &=16\hat i-16\hat j-8\hat k \end{aligned}(4,4,0)×(2,0,4)​=​i^42​j^​40​k^04​​=i^(4⋅4−0⋅0)−j^​(4⋅4−0⋅2)+k^(4⋅0−4⋅2)=16i^−16j^​−8k^​

Thus one perpendicular vector is 16i^−16j^−8k^=8(2i^−2j^−k^)16\hat i-16\hat j-8\hat k=8(2\hat i-2\hat j-\hat k)16i^−16j^​−8k^=8(2i^−2j^​−k^)

  1. Its magnitude is:
162+(−16)2+(−8)2=256+256+64=576=24\sqrt{16^2+(-16)^2+(-8)^2} =\sqrt{256+256+64} =\sqrt{576}=24162+(−16)2+(−8)2​=256+256+64​=576​=24

But we need magnitude 121212, which is half of this. So divide the vector by 222: 8i^−8j^−4k^=4(2i^−2j^−k^)8\hat i-8\hat j-4\hat k=4(2\hat i-2\hat j-\hat k)8i^−8j^​−4k^=4(2i^−2j^​−k^)

  1. Check magnitude:
∣4(2i^−2j^−k^)∣=422+(−2)2+(−1)2=49=12\left|4(2\hat i-2\hat j-\hat k)\right| =4\sqrt{2^2+(-2)^2+(-1)^2} =4\sqrt{9}=12​4(2i^−2j^​−k^)​=422+(−2)2+(−1)2​=49​=12

Hence one such vector is 4(2i^−2j^−k^)4(2\hat i-2\hat j-\hat k)4(2i^−2j^​−k^)

  1. Compare with options:
  • Option A: 4(2i^−2j^−k^)4(2\hat i-2\hat j-\hat k)4(2i^−2j^​−k^) ✅
  • Other options are not scalar multiples of the required perpendicular direction.

Therefore, the correct answer is A.

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