Q.45. Find a unit vector parallel to the xy plane and perpendicular to $4 \hat{i} - 3 \hat{j} + \hat{k}$ .

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Solution

Dear student
$Since the vector is parallel to the xy - plane, its z - component is zero . So the vector is of the form ai + bj . As it is perpendicular to 4 i - 3 j + k, the dot product of the two is zero . This gives a condition on the values of a and b relative to one another . (ai + bj) . (4 i - 3 j + k) = 0 \Rightarrow 4 a - 3 b = 0 . So a = \frac{3 b}{4} . There are infinitely many such vectors, but you are after a unit vector . There will only be two of these . Being a unit vector, we know that a and b must satisfy a^{2} + b^{2} = 1 . Since a = \frac{3 b}{4}, we can substitute to find values of a and b to answer the question . {(\frac{3 b}{4})}^{2} + b^{2} = 1 \Rightarrow (\frac{9}{16} + 1) b^{2} = 1 \Rightarrow b^{2} = \frac{16}{25} . So b = \frac{4}{5} or b = - \frac{4}{5} The corresponding a values are a = \frac{3}{5} or a = - \frac{3}{5} This gives two solutions to choose from \frac{3}{5} i + \frac{4}{5} j and - \frac{3}{5} i - \frac{4}{5} j You can easily double check to see that both of these satisfy all necessary requirements .$
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4 \hat{i} - \hat{j} + 3 \hat{k} and - 2 \hat{i} + \hat{j} - 2 \hat{k} .

\vec{a} = 2 \hat{i} + \hat{j} + \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} .

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