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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Let \(\overrightarrow a = \hat i + 4 \hat j + 2\hat k, \overrightarrow b = 3\hat i − 2 \hat j + 7\hat k \: and \: \overrightarrow c = 2\hat i − \hat j + 4\hat k \) . Find a vector \( \overrightarrow d\) which is perpendicular to both \(\overrightarrow a\) and \(\overrightarrow b\) , and \( \overrightarrow c ⋅ \overrightarrow d =15\)

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1 Answer

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Toolbox:
  • If $\overrightarrow a+\overrightarrow b$ then $\overrightarrow a.\overrightarrow b=0$
  • $(a_1\hat i+a_2\hat j+a_3\hat k).(b_1\hat i+b_2\hat j+b_3\hat k)=a_1b_1+a_2b_2+a_3b_3$
Step 1:
Let $\overrightarrow a=\hat i+4\hat j+2\hat k$
$\overrightarrow b=3\hat i-2\hat j+7\hat k$
$\overrightarrow c=2\hat i-\hat j+4\hat k$
Let $\overrightarrow d$ be $d_1\hat i+d_2\hat j+d_3\hat k$
It is given $\overrightarrow d$ is $\perp$ to vector $\overrightarrow a=\hat i+4\hat j+2\hat k$
$\Rightarrow \overrightarrow a.\overrightarrow d=0$
$d_1+4d_2+2d_3=0$-----(1)
Also vector $\overrightarrow d$ is $\perp$ to vector $\overrightarrow b=3\hat i-2\hat j+7\hat k$
$\Rightarrow \overrightarrow b.\overrightarrow d=0$
$3d_1-2d_2+7d_3=0$-----(2)
Step 2:
Now equating equ(1) and equ(2) we get
$\large\frac{d_1}{\begin{vmatrix}4 & 2\\-2 & 7\end{vmatrix}}=\large\frac{d_2}{\begin{vmatrix}2 & 1\\7 & 3\end{vmatrix}}=\large\frac{d_3}{\begin{vmatrix}1 & 4\\3 & -2\end{vmatrix}}$
$\Rightarrow \large\frac{d_1}{28+4}=\large\frac{d_2}{-1}=\large\frac{d_3}{-14}=$$p$(assume)
Therefore $\large\frac{d_1}{32}=\large\frac{d_2}{-1}=\large\frac{d_3}{-14}=$$p$
(i.e)$d_1=32p:d_2=-p:d_3=-14p$.
Therefore $\overrightarrow d=32p\hat i-p\hat j-14p\hat k$
$\overrightarrow c=2\hat i-p\hat j+4\hat k$
Step 3:
It is given $\overrightarrow c.\overrightarrow d=15$
Therefore $(2\hat i-\hat j+4\hat k).(32p\hat i-p\hat j-14p\hat k)=15$
$\Rightarrow 64p+p-56p=15$
$9p=15$
$p=\large\frac{15}{9}=\frac{5}{3}$
Step 4:
Therefore vector $\overrightarrow d=32p\hat i-p\hat j-14p\hat k$
We know $p=\large\frac{5}{3}$
Substitute the value of p we get
$\overrightarrow d=32\times \large\frac{5}{3}$$\hat i-\large\frac{5}{3}$$\hat j-14\times\large\frac{5}{3}$$\hat k$
$ \overrightarrow d=\large\frac{160}{3}$$\hat i-\large\frac{5}{3}$$\hat j+\large\frac{70}{3}$$\hat k$
answered May 24, 2013 by sreemathi.v
 

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