The volume of the tetrahedron having the edges $\widehat {{\bf{\dot i}}} + 2\,\widehat {{\bf{\dot j}}} - {\bf{\hat k}},\,\;\;\widehat {{\bf{\dot i}}} + \widehat {{\bf{\dot j}}} + {\bf{\hat k}},\widehat {\,{\bf{\dot i}}} - \widehat {{\bf{\dot j}}} + \lambda \,{\bf{\hat k}}$ as coterminous is $\frac{2}{3}$ cu unit. Then, ${\rm{\lambda }}$ equals

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Application +4 | -1

The volume of the tetrahedron having the edges

\widehat {\mathbf{\dot i}} + 2\,\widehat {\mathbf{\dot j}} - \mathbf{\hat k},\,\;\;\widehat {\mathbf{\dot i}} + \widehat {\mathbf{\dot j}} + \mathbf{\hat k},\widehat {\,\mathbf{\dot i}} - \widehat {\mathbf{\dot j}} + \lambda \,\mathbf{\hat k}

as coterminous is

\frac{2}{3}

cu unit. Then,

\mathrm{\lambda }

equals

View answer and explanation

Correct option: A

The correct answer is A (1) because the volume of the tetrahedron is given by one-sixth of the scalar triple product of the coterminous edges, and solving for lambda gives 1.

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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,

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\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}