Two adjacent sides of a parallelogram ABCD is given by \[\overrightarrow {AB} = 2\hat i + 10\hat j + 11\hat k\] and \[\overrightarrow {AD} = - \hat i + 2\hat j + 2\hat k\]. The side AD is rotated by an acute angle \[\alpha \] in the plane of the parallelogram so that AD becomes AD’. If AD’ makes a right angle with the side AB, then what is the cosine of the angle \[\alpha \] ?
(a). \[\dfrac{8}{9}\]
(b). \[\dfrac{{\sqrt {17} }}{9}\]
(c). \[\dfrac{1}{9}\]
(d). \[\dfrac{{4\sqrt 5 }}{9}\]
VerifiedHint: Find the cosine of the angle between the vectors AB and AD using the formula \[\cos \theta = \dfrac{{\overrightarrow {AB} .\overrightarrow {AD} }}{{|\overrightarrow {AB} ||\overrightarrow {AD} |}}\]. This angle can be written as 90° – \[\alpha \]. Then find the value of \[\sin \alpha \] and then \[\cos \alpha \].
The angle between two vectors a and b is given by the formula as follows:
$\cos \theta = \dfrac{{\overrightarrow a .\overrightarrow b }}{{|\overrightarrow a||\overrightarrow b |}}$
The angle between the vectors AB and AD is given as follows:
$\cos \theta = \dfrac{{\overrightarrow {AB} .\overrightarrow {AD} }}{{|\overrightarrow {AB}
||\overrightarrow {AD} |}}$
The vectors are given by \[\overrightarrow {AB} = 2\hat i + 10\hat j + 11\hat k\] and
\[\overrightarrow {AD} = - \hat i + 2\hat j + 2\hat k\], hence, we have:
$\cos \theta = \dfrac{{(2\hat i + 10\hat j + 11\hat k).( - \hat i + 2\hat j + 2\hat k)}}{{\sqrt
{{2^2} + {{10}^2} + {{11}^2}} \sqrt {{{( - 1)}^2} + {2^2} + {2^2}} }}$
Simplifying, we have:
\[\cos \theta = \dfrac{{2( - 1) + 10.2 + 11.2}}{{\sqrt {4 + 100 + 121} \sqrt {1 + 4 + 4} }}\]
Multiplying and adding, we have:
\[\cos \theta = \dfrac{{ - 2 + 20 + 22}}{{\sqrt {225} \sqrt 9 }}\]
We know that the square root of 225 is 15 and the square root of 9 is 3.
\[\cos \theta = \dfrac{{40}}{{15.3}}\]
Simplifying, we have:
\[\cos \theta = \dfrac{8}{9}............(1)\]
From the figure, we know that \[\theta \] is \[90^\circ - \alpha \].
Then, in equation (1), we have:
\[\cos (90^\circ - \alpha ) = \dfrac{8}{9}\]
We know that the value of \[\cos (90^\circ - x)\] is \[\sin x\], then, we have:
\[\sin \alpha = \dfrac{8}{9}............(2)\]
We know that \[{\cos ^2}x = 1 - {\sin ^2}x\], then using equation (2), we can find the value of \[\cos \alpha \].
\[{\cos ^2}\alpha = 1 - {\sin ^2}\alpha \]
\[{\cos ^2}\alpha = 1 - {\left( {\dfrac{8}{9}} \right)^2}\]
Simplifying, we have:
\[{\cos ^2}\alpha = 1 - \dfrac{{64}}{{81}}\]
Taking common denominator, we have:
\[{\cos ^2}\alpha = \dfrac{{81 - 64}}{{81}}\]
\[{\cos ^2}\alpha = \dfrac{{17}}{{81}}\]
Taking square root and retaining only the positive value, we have:
\[\cos \alpha = \sqrt {\dfrac{{17}}{{81}}} \]
Simplifying, we have:
\[\cos \alpha = \dfrac{{\sqrt {17} }}{9}\]
Hence, the correct answer is option (b).
Note: Always choose positive or negative root by the value of the trigonometric functions when the angle is in one of the four quadrants.
