JEE MATHEMATICS MCQ: CROSS PRODUCT

1 Easy

1 of 1

Understanding +4 | -1

Consider a tetrahedron with faces

{F}_1, {~F}_2, {~F}_3, {~F}_4

. Let

{V}_1, {V}_2, {V}_3, {V}_4

be the vectors whose magnitudes are respectively equal to areas of

{F}_1, {~F}_2, {~F}_3, {~F}_4

and whose directions are perpendicular to their faces in outward direction. Then

\left|{V}_1+{V}_2+{V}_3+{V}_4\right|

equals

View answer and explanation

Correct option: C

Option C (0) is correct because the sum of the vectors representing the areas of the faces of a tetrahedron, directed outward, is zero.

More Questions from CROSS PRODUCT

Practice All Chapter MCQs →

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

\vec{a}, \vec{b}

\lambda

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

x.

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

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

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

a \times b

c

30^\circ ,

|(a \times b) \times c|

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

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

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

O

S

O A

O C

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