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

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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 $\overrightarrow a=3\hat i+2\hat j+2\hat k$ and $\overrightarrow b=\hat i+2\hat j-2\hat k$
Let us find $\overrightarrow a+\overrightarrow b=(3\hat i+2\hat j+2\hat k)+(\hat i+2\hat j-2\hat k)$
$\qquad\qquad\qquad\quad=4\hat i+4\hat j+0\hat k$
$\qquad\qquad\qquad\quad=4\hat i+4\hat j$
Step 2:
Next let us find $\overrightarrow a-\overrightarrow b$
$\overrightarrow a-\overrightarrow b=(3\hat i+2\hat j+2\hat k)-(\hat i+2\hat j-2\hat k)$
$\qquad\quad=2\hat i-0\hat j+4\hat k$
$\qquad\quad=2\hat i+4\hat k$
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 $(\overrightarrow a-\overrightarrow b)$
$\hat n=\pm\large\frac{(4\hat i+4\hat j)\times(2\hat i+4\hat k)}{|4\hat i+4\hat j\times 2\hat i+4\hat k|}$
Step 4:
Let us determine $(4\hat i+4\hat j)\times (2\hat i+4\hat k)$
$(4\hat i+4\hat j)\times (2\hat i+4\hat k)=\begin{vmatrix}i & j & k\\4 & 4 & 0\\2 & 0 & 4\end{vmatrix}$
$\qquad\qquad\qquad\qquad=\hat i(16-0)-\hat j(16-0)+\hat k(0-8)$
$\qquad\qquad\qquad\qquad=16\hat i-16\hat j-8\hat k$
Step 5:
$|(4\hat i+4\hat j)\times (2\hat i+4\hat k)|=\sqrt{16^2+(-16)^2+(-8)^2}$
$\qquad\qquad\qquad\qquad\quad=\sqrt{256+256+64}$
$\qquad\qquad\qquad\qquad\quad=\sqrt{576}$
$\qquad\qquad\qquad\qquad\quad=24.$
Step 6:
Therefore the required unit vector $\hat n=\pm\large\frac{16\hat i-16\hat j-8\hat k}{24}$
$\qquad\qquad\qquad\qquad\qquad\qquad\quad=\pm\big(\large\frac{16}{24}$$\hat i-\large\frac{16}{24}$$\hat j-\large\frac{8}{24}$$\hat k\big)$
$\qquad\qquad\qquad\qquad\qquad\qquad\quad=\pm\big(\large\frac{2}{3}$$\hat i-\large\frac{2}{3}$$\hat j-\large\frac{2}{3}$$\hat k\big)$
answered May 22, 2013 by sreemathi.v
 

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