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Q12 Let $\vec a = \hat i + 4 \hat j + 2 \hat k , \vec b = 3 \hat i - 2 \hat j + 7 \hat k \: \:and \: \: \vec c = 2 \hat i - \hat j + 4 \hat k$ . Find a vector $\vec d$ which is perpendicular to both $\vec a \: \: and \: \: \vec b \: \: and \: \: \vec c . \vec d = 15$

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Given,

$\vec a = \hat i + 4 \hat j + 2 \hat k , \vec b = 3 \hat i - 2 \hat j + 7 \hat k \: \:and \: \: \vec c = 2 \hat i - \hat j + 4 \hat k$

Let $\vec d=d_1\hat i+d_2\hat j +d_3\hat k$

now, since it is given that d is perpendicular to $\vec a$ and $\vec b$ , we got the condition,

$\vec b.\vec d=0$ and $\vec a.\vec d=0$

$(\hat i+4\hat j +2\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=0$ And $(3\hat i-2\hat j +7\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=0$

$d_1+4d_2+2d_3=0$ And $3d_1-2d_2+7d_3=0$

here we got 2 equation and 3 variable. one more equation will come from the condition:

$\vec c . \vec d = 15$

$(2\hat i-\hat j +4\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=15$

$2d_1-d_2+4d_3=15$

so now we have three equation and three variable,

$d_1+4d_2+2d_3=0$

$3d_1-2d_2+7d_3=0$

$2d_1-d_2+4d_3=15$

On solving this three equation we get,

$d_1=\frac{160}{3},d_2=-\frac{5}{3}\:and\:d_3=-\frac{70}{3}$ ,

Hence Required vector :

$\vec d=\frac{160}{3}\hat i-\frac{5}{3}\hat j-\frac{70}{3}\hat k$ .

Posted by

Pankaj Sanodiya

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