Question

If \[\vec a = \hat i + 2\hat j + 2\hat k\] , \[\left| {\vec b} \right| = 5\] and the angle between \[\vec a\] and \[\vec b\] is\[\dfrac{\pi }{6}\] , then the area of the triangle formed by these two vectors are two sides is :
A.\[\dfrac{{15}}{4}\]
B.\[\dfrac{{15}}{2}\]
C.\[15\]
D.\[\dfrac{{15\sqrt 3 }}{2}\]
E.\[15\sqrt 3 \]

Answer 1

Hint: We have given a vector \[\vec a = \hat i + 2\hat j + 2\hat k\] with coordinates and other vector \[\vec b\] with magnitude so , for the area of the triangle we have a formula as \[ = \dfrac{1}{2}\left| {\vec a \times \vec b} \right|\] , which half times the cross product of both the vectors . So , now we can use the given magnitude of the vector \[\vec b\] .

Complete step-by-step answer:
Given : \[\vec a = i + 2j + 2k\] , \[\left| {\vec b} \right| = 5\] and angle between them \[ = \dfrac{\pi }{6}\]
Now , using the formula for area of the triangle formed by the two vectors is \[ = \dfrac{1}{2}\left| {\vec a \times \vec b} \right|\] .
On simplifying the cross product we get ,
\[ = \dfrac{1}{2}\left| a \right|\left| b \right|\sin \theta \] , where \[\theta \] is the angle between the vectors \[\vec a\] and \[\vec b\] .
Now , for the magnitude of \[\vec a\] we use formula
\[ = \sqrt {{x^2} + {y^2} + {z^2}} \] , where \[x,y,z\] are coefficient of the unit vectors \[\hat i,\hat j,\hat k\] respectively .
Putting the values of \[x,y,z\] for vector \[\vec a\] we get ,
\[ = \sqrt {{1^2} + {2^2} + {2^2}} \] , on simplifying we get
\[ = \sqrt 9 \] , on solving we get
\[ = 3\] .
Therefore , \[\left| {\vec a} \right| = 3\]
Now putting the values in the above formula \[ = \dfrac{1}{2}\left| a \right|\left| b \right|\sin \theta \] , we get .
\[ = \dfrac{1}{2} \times 3 \times 5 \times \left( {\sin \dfrac{\pi }{6}} \right)\]
Putting the value of \[\sin \dfrac{\pi }{6} = \dfrac{1}{2}\] , we get
\[ = \dfrac{1}{2} \times 15 \times \dfrac{1}{2}\]
On solving we get
\[ = \dfrac{{15}}{4}\] .
Therefore , option (A ) is the correct answer .
So, the correct answer is “Option A”.

Note: You have to remember the formula for the dot product or the cross product of two vectors , in terms of their magnitude and angle between them . In vector theory, vectors are visualized as directed line segments whose lengths are equal to their magnitudes . The formula for the area of a triangle is derived from laws of parallelogram of vectors . The magnitude of the cross product of two vectors equals the area of the parallelogram spanned by these two vectors . And since a triangle is half of a parallelogram .