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MHTCET 11 May 2024 Shift 1 Paper

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Question : 136 of 150

Marks: +1, -0

The area of the triangle with vertices

(1,2,0),(1,0,2)

(0,3,1)

is

$\sqrt{3}$ sq. units
$\sqrt{6}$ sq. units
$\sqrt{5}$ sq. units
$\sqrt{7}^{\text {sq. units }}$

Solution:

Concept:

The area of a triangle in space is given by

\frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|

.

Explanation:

Let

A(1,2,0)

,

B(1,0,2)

, and

C(0,3,1)

.

\overrightarrow{AB} = B - A = (0,-2,2)

.

\overrightarrow{AC} = C - A = (-1,1,1)

.

Now compute the cross product:

\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -2 & 2 \\ -1 & 1 & 1 \end{vmatrix}

= \hat{i}(-2-2) - \hat{j}(0+2) + \hat{k}(0-2)

= -4\hat{i} - 2\hat{j} - 2\hat{k}

.

So,

|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{16+4+4} = \sqrt{24} = 2\sqrt{6}

.

Therefore, area of the triangle

= \frac{1}{2}(2\sqrt{6}) = \sqrt{6}

sq. units.

Answer:

\sqrt{6}

sq. units (Option B).

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