If ${\bf{\vec a}},\;{\bf{\vec b}},\;{\bf{\vec c}}$ are non-coplanar vectors and $\left( {{\bf{\vec a}} - {\rm{\lambda }}\;{\bf{\vec b}}} \right)\cdot\left( {{\bf{\vec b}} - 2{\bf{\vec c}}} \right) \times \left( {{\bf{\vec c}} + 2{\bf{\vec a}}} \right) = 0,$ then λ is equal to

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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Analyzing +4 | -1

\mathbf{\vec a},\;\mathbf{\vec b},\;\mathbf{\vec c}

are non-coplanar vectors and

\left( {\mathbf{\vec a} - \mathrm{\lambda }\;\mathbf{\vec b}} \right)\cdot\left( {\mathbf{\vec b} - 2\mathbf{\vec c}} \right) \times \left( {\mathbf{\vec c} + 2\mathbf{\vec a}} \right) = 0,

then λ is equal to

View answer and explanation

Correct option: D

- 1/4

The correct answer is D (

- 1/4

) because the scalar triple product is zero, implying coplanarity, which leads to the solution λ = -1/4.

Practice All Chapter MCQs →

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

\vec r,

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

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

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

\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}

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

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

\vec{a}, \vec{b}

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

a \cdot(a \times b)=

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