Let a = 2 i + 3 j + 3 k and b = 6 i + 3 j + 3 k . Then the square...

Let

\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}

and

\vec{b} = 6\hat{i} + 3\hat{j} + 3\hat{k}

. Then the square of the area of the triangle with adjacent sides determined by the vectors

(2\vec{a} + 3\vec{b})

and

(\vec{a} - \vec{b})

is :

Options

A $450$
B $900$
C $1800$
D $2400$

Correct answer

C. $1800$

Step-by-step solution

The area of a triangle with adjacent sides

\vec{u}

and

\vec{v}

is given by

\Delta = \dfrac{1}{2} |\vec{u} \times \vec{v}|

.

Here,

\vec{u} = 2\vec{a} + 3\vec{b}

and

\vec{v} = \vec{a} - \vec{b}

\vec{u} \times \vec{v} = (2\vec{a} + 3\vec{b}) \times (\vec{a} - \vec{b})

\vec{u} \times \vec{v} = 2(\vec{a} \times \vec{a}) - 2(\vec{a} \times \vec{b}) + 3(\vec{b} \times \vec{a}) - 3(\vec{b} \times \vec{b})

Since

\vec{a} \times \vec{a} = \vec{0}

\vec{b} \times \vec{b} = \vec{0}

, and

\vec{b} \times \vec{a} = -\vec{a} \times \vec{b}

, we get:

\vec{u} \times \vec{v} = -2(\vec{a} \times \vec{b}) - 3(\vec{a} \times \vec{b}) = -5(\vec{a} \times \vec{b})

The area of the triangle is

\Delta = \dfrac{1}{2} |-5(\vec{a} \times \vec{b})| = \dfrac{5}{2} |\vec{a} \times \vec{b}|

.

Now, calculating

\vec{a} \times \vec{b}

\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 3 \\ 6 & 3 & 3 \end{vmatrix}

\vec{a} \times \vec{b} = \hat{i}(9 - 9) - \hat{j}(6 - 18) + \hat{k}(6 - 18) = 12\hat{j} - 12\hat{k}

The square of the magnitude is:

|\vec{a} \times \vec{b}|^2 = 12^2 + (-12)^2 = 144 + 144 = 288

The square of the area of the triangle is:

\Delta^2 = \left( \dfrac{5}{2} \right)^2 |\vec{a} \times \vec{b}|^2 = \dfrac{25}{4} \times 288 = 25 \times 72 = 1800

Answer:

1800

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