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Vector Algebra question

2003 · Shift 0 · Q90
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Vector Algebra question

2003 · Shift 0 · Q90

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let u→=i^+j^, v→=i^−j^\overrightarrow u = \widehat i + \widehat j,\,\overrightarrow v = \widehat i - \widehat ju=i+j​,v=i−j​ and w→=i^+2j^+3k^  .\overrightarrow w = \widehat i + 2\widehat j + 3\widehat k\,\,.w=i+2j​+3k. If n^\widehat nn is a unit vector such that u→.n^=0\overrightarrow u .\widehat n = 0u.n=0 and v→.n^=0  ,\overrightarrow v .\widehat n = 0\,\,,v.n=0, then ∣w→.n^∣\left| {\overrightarrow w .\widehat n} \right|​w.n​ is equal to :
  1. A
    333
  2. B
    000
  3. C
    111
  4. D
    222
View written solutionFree

Correct answer: A

  1. Write the given vectors in component form:
u⃗=(1,1,0),v⃗=(1,−1,0),w⃗=(1,2,3)\vec u = (1,1,0), \quad \vec v = (1,-1,0), \quad \vec w = (1,2,3)u=(1,1,0),v=(1,−1,0),w=(1,2,3)

We need a unit vector n^=(x,y,z)\hat n=(x,y,z)n^=(x,y,z) such that

u⃗⋅n^=0,v⃗⋅n^=0.\vec u \cdot \hat n = 0, \qquad \vec v \cdot \hat n = 0.u⋅n^=0,v⋅n^=0.
  1. Apply the orthogonality conditions:

From u⃗⋅n^=0\vec u \cdot \hat n=0u⋅n^=0,

x+y=0x+y=0x+y=0

From v⃗⋅n^=0\vec v \cdot \hat n=0v⋅n^=0,

x−y=0x-y=0x−y=0
  1. Solve these equations:

Adding both equations,

2x=0  ⟹  x=02x=0 \implies x=02x=0⟹x=0

Then from x+y=0x+y=0x+y=0,

y=0y=0y=0

So,

n^=(0,0,z)\hat n=(0,0,z)n^=(0,0,z)

Since n^\hat nn^ is a unit vector,

∣n^∣=1  ⟹  z=±1|\hat n|=1 \implies z=\pm 1∣n^∣=1⟹z=±1

Hence,

n^=±k^\hat n=\pm \hat kn^=±k^
  1. Now compute w⃗⋅n^\vec w\cdot \hat nw⋅n^:
w⃗⋅n^=(1,2,3)⋅(0,0,±1)=±3\vec w\cdot \hat n = (1,2,3)\cdot(0,0,\pm 1)=\pm 3w⋅n^=(1,2,3)⋅(0,0,±1)=±3

Therefore,

∣w⃗⋅n^∣=3\left|\vec w\cdot \hat n\right|=3∣w⋅n^∣=3
  1. Check options:
  • A: 333 ✅
  • B: 000 ❌
  • C: 111 ❌
  • D: 222 ❌

So the correct answer is A.

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