The unit vector which is orthogonal to the vector $3\widehat{i} + 2\widehat{j} + 6\widehat{k}$ and is coplanar with the vectors $ 2\widehat{i} + \widehat{j} + \widehat{k}$ and $\widehat{i} + \widehat{j} + \widehat{k}$ is
- $\frac{2\widehat{i}\, - \, 6\widehat{j}\, +\, \widehat{k}}{\sqrt {41}} $
- $\frac{2\widehat{i}\, - \, 3\widehat{j}\, }{\sqrt {13}} $
- $\frac{3\widehat{i}\, -\, \widehat{k}\, }{\sqrt {10}} $
- $\frac{4\widehat{i}\, + \, 3\widehat{j}\, -\, 3\widehat{k}}{\sqrt {34}} $
The Correct Option is C
Solution and Explanation
orthogonal to $ \overrightarrow {c}$ is $\lambda \{(\overrightarrow {a} \times \overrightarrow {b}) \times \overrightarrow {c} \}.$
$\therefore$ A vector coplanar to $ (2\widehat{i} + \widehat{j} + \widehat{k}), (\widehat{i} + \widehat{j} + \widehat{k})$ and
orthogonal to $3\widehat{i} + 2\widehat{j} + 6\widehat{k}$
$ \hspace10mm$ $=\lambda \, [\{(2\widehat{i} + \widehat{j} + \widehat{k}) \times (\widehat{i} + \widehat{j} + \widehat{k})\} \, \, \times (3\widehat{i} + 2\widehat{j} + 6\widehat{k})]$
$ \hspace10mm$ $=\lambda \, [(2\widehat{i} + \widehat{j} + 3\widehat{k}) \times (3\widehat{i} + 2\widehat{j} + 6\widehat{k})] = \lambda( 21\widehat{j} - 7\widehat{k})$
$\therefore$ Unit vector $= +\frac{( 21\widehat{j} - 7\widehat{k})}{\sqrt {(21)^2+(7)^2}} \, = + \frac{(3\widehat{i} - \widehat{k})}{\sqrt 10} $
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Concepts Used:
Vector Algebra
A vector is an object which has both magnitudes and direction. It is usually represented by an arrow which shows the direction(→) and its length shows the magnitude. The arrow which indicates the vector has an arrowhead and its opposite end is the tail. It is denoted as
The magnitude of the vector is represented as |V|. Two vectors are said to be equal if they have equal magnitudes and equal direction.
Vector Algebra Operations:
Arithmetic operations such as addition, subtraction, multiplication on vectors. However, in the case of multiplication, vectors have two terminologies, such as dot product and cross product.