JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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Let area of faces,

\triangle O A B=\lambda_1, \Delta O A C=\lambda_2, \Delta O B C=\lambda_3, \Delta A B C=\lambda_4

and

h_1, h_2, h_3, h_4

be perpendicular height from 0 to face

\triangle A B C, A

to the face

\triangle O B C, B

to the face

\triangle O A C, C

to the face

\triangle O A B

, then the face

\frac{1}{3} \lambda_1 h_4 \cdot \frac{1}{3} \lambda_2 h_3+\frac{1}{3} \lambda_3 h_2+\frac{1}{3} \lambda_4 h_1

View answer and explanation

Correct option: A

\frac{2}{3}|[AB\,\,AC\,\,OA]|

Option A is correct because the sum of the volumes of the tetrahedra formed by dropping perpendiculars from the origin to each face equals twice the volume of the tetrahedron.

More Questions from SCALAR TRIPLE PRODUCT

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\vec r\cdot\vec a = \vec r\cdot\vec b = \vec r\cdot\vec c = 0

for some non-zero vector

\vec r,

then the value of

\left[ {\vec a\vec b\vec c} \right],

is The shortest distance between the lines

\frac{{x - 3}}{2} = \frac{{y - 2}}{3} = \frac{{z - 1}}{{ - 1}}

and

\frac{{x + 3}}{2} = \frac{{y - 6}}{1} = \frac{{z - 5}}{3},

is Let

\mathbf{\vec a} = \widehat \mathbf{i} - \mathbf{\hat k},\;\mathbf{\vec b} = x\widehat \mathbf{i} + \widehat \mathbf{j} + \left( {1 - x} \right)\mathbf{\hat k}

and

\mathbf{\vec c} = y\widehat \mathbf{i} + x\widehat \mathbf{j} + \left( {1 + x - y} \right)\mathbf{\hat k}.

Then

\left[ {\mathbf{\vec a},\;\mathbf{\vec b},\;\mathbf{\vec c}} \right]

depends on If

\vec{a}, \vec{b}

are non-zero vectors, then

((\vec{a} \times \vec{b}) \times \vec{a}) \cdot((\vec{b} \times \vec{a}) \times \vec{b})

equals

a \cdot(a \times b)=

The length of the perpendicular from the origin to the plane passing through the point a and containing the line

\mathbf{r}=\mathbf{b}+\lambda \mathbf{c}