$\overrightarrow {{\bf{AB}}} \times \overrightarrow {{\bf{AC}}} = 2\widehat {\bf{i}} - 4\widehat {\bf{j}} + 4{\bf{\hat k}},$ then the area of $\Delta \;ABC$ is

JEE MATHEMATICS MCQ: CROSS PRODUCT

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Application +4 | -1

\overrightarrow {\mathbf{AB}} \times \overrightarrow {\mathbf{AC}} = 2\widehat \mathbf{i} - 4\widehat \mathbf{j} + 4\mathbf{\hat k},

then the area of

\Delta \;ABC

View answer and explanation

Correct option: A

3 sq units

Option A (3 sq units) is correct because the area of the triangle is half the magnitude of the cross product of the vectors representing two of its sides.

More Questions from CROSS PRODUCT

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A solution of the vector equation

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

, where

\vec{a}, \vec{b}

are two given vectors, is where

\lambda

is a parameter. Let

a = 3\hat i + 2\hat j + x\hat k\mathrm{ }\,\mathrm{\,\,and}\,\,\mathrm{ }b = \hat i - \hat j + \hat k

for some real

x.

Then

\left| {a \times b} \right| = r

is possible if A tetrahedron has vertices P(1, 2, 1), Q(2, 1, 3), R(-1, 1, 2) and O(0, 0, 0). The angle between the faces OPQ and PQR is Let

a = 2\widehat i + \widehat j - 2\widehat k{\text{ }}\,\,{\text{and }}\,\,b = \widehat i + \widehat j.

If c is a vector such that

a \cdot c = |c|,\,\,|c - a\mid = 2\sqrt 2

and the angle between

a \times b

and

c

30^\circ ,

then

|(a \times b) \times c|

equal to If

\vec{A}=i-2 j-3 k, \vec{B}=2 i+j-k, \vec{C}=i+3 j-2 k

, then

(\vec{A} \times \vec{B}) \times \vec{C}

is Let

\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=12 \vec{a}+4 \vec{b}

, where

O

is the origin, If

S

is the parallelogram with adjacent sides

O A

and

O C

, then

\frac{\text { area of the quadrilateral } O A B C}{\text { area of } S}

is equal to