The value of $\lambda $, for which the four points $2\,{\bf{\hat i}} + 3\,{\bf{\hat j}} - {\bf{\hat k}},\,\;{\bf{\hat i}} + 2\,{\bf{\hat j}} + 3\,{\bf{\hat k}},{\rm{\;}}3\,{\bf{\hat i}} + 4\,{\bf{\hat j}} - 2\,{\bf{\hat k}},{\rm{\;}}\,{\bf{\hat i}} - {\rm{\lambda }}\,{\bf{\hat j}} + 6\,{\bf{\hat k}}$ are coplanar, is

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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Application +4 | -1

The value of

\lambda

, for which the four points

2\,\mathbf{\hat i} + 3\,\mathbf{\hat j} - \mathbf{\hat k},\,\;\mathbf{\hat i} + 2\,\mathbf{\hat j} + 3\,\mathbf{\hat k},\mathrm{\;}3\,\mathbf{\hat i} + 4\,\mathbf{\hat j} - 2\,\mathbf{\hat k},\mathrm{\;}\,\mathbf{\hat i} - \mathrm{\lambda }\,\mathbf{\hat j} + 6\,\mathbf{\hat k}

are coplanar, is

View answer and explanation

Correct option: A

- 2

Option A (-2) is correct because the coplanarity condition requires the scalar triple product of the vectors formed by the points to be zero, which leads to an equation that can be solved for

\lambda

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}