JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

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Analyzing +4 | -1

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

are non-coplanar vectors and

\lambda

is a real number, then

\left[\begin{array}{lll}\lambda(\mathbf{a}+\mathbf{b}) & \lambda^2 \mathbf{b} & \lambda \mathbf{c}\end{array}\right]=\left[\begin{array}{l}\mathbf{a} \mathbf{b}+\mathbf{c} \mathbf{b}\end{array}\right]

for

View answer and explanation

Correct option: C

no value of

\lambda

Option C (no value of

\lambda

) is correct because the equation simplifies to

\lambda^4 = -1

, which has no real solutions for

\lambda

More Questions from SCALAR TRIPLE PRODUCT

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}