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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find a unit vector perpendicular to each of the vector \( \overrightarrow a + \overrightarrow b\) and \( \overrightarrow a - \overrightarrow b\), where \( \overrightarrow a = \hat i + \hat j + \hat k\) and \( \overrightarrow b = \hat i + 2\hat j +3\hat k\)

$(a)\;\pm\large\frac{-2\hat i+4\hat j-2\hat k}{\sqrt{24}}\qquad(b)\;\pm\large\frac{2\hat i+4\hat j+2\hat k}{\sqrt{24}}\qquad(c)\;\pm\large\frac{-2\hat i-4\hat j-2\hat k}{\sqrt{24}}\qquad(d)\;\pm\large\frac{2\hat i+4\hat j-2\hat k}{24}$

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1 Answer

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Toolbox:
  • Unit vector $\perp$ to two vectors $\overrightarrow a$ and $\overrightarrow b$ is given by $\hat n=\pm\large\frac{\overrightarrow a\times\overrightarrow b}{|\overrightarrow a\times\overrightarrow b|}$
  • $\overrightarrow a\times \overrightarrow b=\hat i(a_2b_3-a_3b_2)-\hat j(a_1b_3-a_3b_1)+\hat k(a_1b_2-a_2b_1)$
  • Where $\overrightarrow a=(a_1\hat i+a_2\hat j+a_3\hat k)$ and $\overrightarrow b=(b_1\hat i+b_2\hat j+b_3\hat k)$
Step 1:
Let $\overrightarrow a=\hat i+\hat j+\hat k$ and $\overrightarrow b=\hat i+2\hat j+3\hat k$
Let us find $\overrightarrow a+\overrightarrow b=(\hat i+\hat j+\hat k)+(\hat i+2\hat j+3\hat k)$
$\qquad\qquad\qquad\quad=2\hat i+3\hat j+4\hat k$
Step 2:
Next let us find $\overrightarrow a-\overrightarrow b$
$\overrightarrow a-\overrightarrow b=(\hat i+\hat j+\hat k)-(\hat i+2\hat j+3\hat k)$
$\qquad\quad=0\hat i-1\hat j-2\hat k$
$\qquad\quad=-\hat j-2\hat k$
Step 3:
Unit vector $\perp$ to $\overrightarrow a$ and $\overrightarrow b$ is given by $\hat n=\pm\large\frac{\overrightarrow a\times\overrightarrow b}{|\overrightarrow a\times\overrightarrow b|}$
Here we are asked to find a unit vector which is $\perp$ to $(\overrightarrow a+\overrightarrow b)$ and $(\overrightarrow a-\overrightarrow b)$
$\hat n=\pm\large\frac{(2\hat i+3\hat j+4\hat k)\times(-\hat j-2\hat k)}{|(2\hat i+3\hat j+4\hat k)\times (-\hat j-2\hat k)|}$
Step 4:
Let us determine $(2\hat i+3\hat j+4\hat k)\times (-\hat j-2\hat k)$
$(2\hat i+3\hat j+4\hat k)\times (-\hat j-2\hat k)=\begin{vmatrix}i & j & k\\2 & 3 & 4\\0 & -1 & -2\end{vmatrix}$
$\qquad\qquad\qquad\qquad=\hat i(-6+4)-\hat j(-4-0)+\hat k(-2-0)$
$\qquad\qquad\qquad\qquad=-2\hat i+4\hat j-2\hat k$
Step 5:
$|(2\hat i+3\hat j+4\hat k)\times (-\hat j-2\hat k)|=\sqrt{(-2)^2+4^2+(-2)^2}$
$\qquad\qquad\qquad\qquad\quad=\sqrt{4+16+4}$
$\qquad\qquad\qquad\qquad\quad=24.$
Step 6:
$\therefore\:$ The required unit vector $\hat n=\pm\large\frac{-2\hat i+4\hat j-2\hat k}{\sqrt{24}}$
answered Mar 21, 2014 by rvidyagovindarajan_1
edited Mar 21, 2014 by rvidyagovindarajan_1
 

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