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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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For any vector $\overrightarrow{a}$,the value of $(\overrightarrow{a}.\hat i)^2+ (\overrightarrow{a}.\hat j)^2+ (\overrightarrow{a}.\hat k)^2 $ is equal to

\[(A)\;{\mid\overrightarrow{a}\mid}^2\quad(B)\;3{\mid\overrightarrow{a}\mid}^2\quad(C)\;4{\mid\overrightarrow{a}\mid}^2\quad(A)\;2{\mid\overrightarrow{a}\mid}^2\]
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1 Answer

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Toolbox:
  • $\hat i \times \hat i=0,\hat j \times \hat j=0,\hat k \times \hat k=0$
  • $\hat i \times \hat j=\hat k,\hat j \times \hat k=\hat i\;and \; \hat k \times \hat i=\hat j$
Given $|\overrightarrow a \times \hat i|^2+|\overrightarrow a \times \hat j|^2+|\overrightarrow a \times \hat k|^2$
Let $\overrightarrow a =a_1\hat i+a_2 \hat j+a_3 \hat k$
Then $\overrightarrow a \times \hat i=(a_1 \hat i+a_2 \hat j+a_3 \hat k) \times \hat i$
On expanding we get,
$a_1(\hat i \times \hat i)+a_2 (\hat j \times \hat i)+a_3 (\hat k \times \hat i)$
We know $\hat i \times \hat i=0,\hat j \times \hat i=-\hat k\;and \; \hat k \times \hat i=\hat j$
$=-a_2 \hat k+a_3 \hat j$
Therefore $|\overrightarrow a \times \hat i|^2={a_2}^2+{a_3}^2$
Similarly $|\overrightarrow a \times \hat j |^2=(a_1 \hat i+a_2 \hat j+a_3 \hat k) \times \hat j$
$=a_1(\hat i \times \hat j)+a_2 (\hat j \times \hat j)+a_3 (\hat k \times \hat j)$
$=a_1 \hat k-a_3 \hat i$
Therefore $|\overrightarrow a \times \hat j|^2={a_1}^2+{a_3}^2$
Similarly $|\overrightarrow a \times \hat k |^2=(a_1 \hat i+a_2 \hat j+a_3 \hat k) \times \hat k$
$=a_1(\hat i \times \hat k)+a_2 (\hat j \times \hat k)+a_3 (\hat k \times \hat k)$
$=-a_1 \hat j+a_2 \hat i$
$|\overrightarrow a \times \hat k|^2={a_1}^2+{a_2}^2$
Therefore $|\overrightarrow a \times \hat i|^2+|\overrightarrow a \times \hat j|^2+|\overrightarrow a \times \hat k|^2$
$={a_2}^2+{a_3}^2+{a_1}^2+{a_3}^2+{a_1}^2+{a_2}^2$
$=2({a_1}^{2}+{a_2}^2+{a_3}^2)$
But ${a_1}^2+{a_2}^2+{a_3}^2=|\overrightarrow a|^2$
$=2 |\overrightarrow a|^2$
Hence $D$ is the correct optiuon
answered May 29, 2013 by meena.p
 

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