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Vector Algebra question

2005 · Shift 0 · Q112
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  5. /2005 · Shift 0 · Q112

Vector Algebra question

2005 · Shift 0 · Q112

JEE MainMathematicsVector AlgebraMCQ+4 / −1
For any vector a→{\overrightarrow a }a, the value of (a→×i^)2+(a→×j^)2+(a→×k^)2{\left( {\overrightarrow a \times \widehat i} \right)^2} + {\left( {\overrightarrow a \times \widehat j} \right)^2} + {\left( {\overrightarrow a \times \widehat k} \right)^2}(a×i)2+(a×j​)2+(a×k)2 is equal to :
  1. A
    3a→23{\overrightarrow a ^2}3a2
  2. B
    a→2{\overrightarrow a ^2}a2
  3. C
    2a→22{\overrightarrow a ^2}2a2
  4. D
    4a→24{\overrightarrow a ^2}4a2
View written solutionFree

Correct answer: C

  1. Let
a⃗=xi^+yj^+zk^\vec a = x\hat i + y\hat j + z\hat ka=xi^+yj^​+zk^

Then

a⃗ 2=x2+y2+z2.\vec a^{\,2} = x^2 + y^2 + z^2.a2=x2+y2+z2.
  1. Compute each cross product.
  • With i^\hat ii^:
a⃗×i^=(xi^+yj^+zk^)×i^\vec a \times \hat i = (x\hat i + y\hat j + z\hat k) \times \hat ia×i^=(xi^+yj^​+zk^)×i^

Using

i^×i^=0,j^×i^=−k^,k^×i^=j^,\hat i \times \hat i = 0,\quad \hat j \times \hat i = -\hat k,\quad \hat k \times \hat i = \hat j,i^×i^=0,j^​×i^=−k^,k^×i^=j^​,

we get

a⃗×i^=y(−k^)+zj^=zj^−yk^.\vec a \times \hat i = y(-\hat k) + z\hat j = z\hat j - y\hat k.a×i^=y(−k^)+zj^​=zj^​−yk^.

Hence,

(a⃗×i^)2=z2+y2.(\vec a \times \hat i)^2 = z^2 + y^2.(a×i^)2=z2+y2.
  • With j^\hat jj^​:
a⃗×j^=x(i^×j^)+y(j^×j^)+z(k^×j^)\vec a \times \hat j = x(\hat i \times \hat j) + y(\hat j \times \hat j) + z(\hat k \times \hat j)a×j^​=x(i^×j^​)+y(j^​×j^​)+z(k^×j^​)

Using

i^×j^=k^,j^×j^=0,k^×j^=−i^,\hat i \times \hat j = \hat k,\quad \hat j \times \hat j = 0,\quad \hat k \times \hat j = -\hat i,i^×j^​=k^,j^​×j^​=0,k^×j^​=−i^,

we get

a⃗×j^=xk^−zi^.\vec a \times \hat j = x\hat k - z\hat i.a×j^​=xk^−zi^.

Hence,

(a⃗×j^)2=x2+z2.(\vec a \times \hat j)^2 = x^2 + z^2.(a×j^​)2=x2+z2.
  • With k^\hat kk^:
a⃗×k^=x(i^×k^)+y(j^×k^)+z(k^×k^)\vec a \times \hat k = x(\hat i \times \hat k) + y(\hat j \times \hat k) + z(\hat k \times \hat k)a×k^=x(i^×k^)+y(j^​×k^)+z(k^×k^)

Using

i^×k^=−j^,j^×k^=i^,k^×k^=0,\hat i \times \hat k = -\hat j,\quad \hat j \times \hat k = \hat i,\quad \hat k \times \hat k = 0,i^×k^=−j^​,j^​×k^=i^,k^×k^=0,

we get

a⃗×k^=−xj^+yi^.\vec a \times \hat k = -x\hat j + y\hat i.a×k^=−xj^​+yi^.

Hence,

(a⃗×k^)2=x2+y2.(\vec a \times \hat k)^2 = x^2 + y^2.(a×k^)2=x2+y2.
  1. Add them:
(a⃗×i^)2+(a⃗×j^)2+(a⃗×k^)2(\vec a \times \hat i)^2 + (\vec a \times \hat j)^2 + (\vec a \times \hat k)^2(a×i^)2+(a×j^​)2+(a×k^)2 =(y2+z2)+(x2+z2)+(x2+y2)= (y^2+z^2) + (x^2+z^2) + (x^2+y^2)=(y2+z2)+(x2+z2)+(x2+y2) =2(x2+y2+z2)= 2(x^2+y^2+z^2)=2(x2+y2+z2) =2a⃗ 2.= 2\vec a^{\,2}.=2a2.
  1. Therefore, the correct option is
C: 2a⃗ 2\boxed{C:~2\vec a^{\,2}}C: 2a2​
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