Let ${\bf{\vec a}} = \widehat {\bf{i}} - {\bf{\hat k}},\;{\bf{\vec b}} = x\widehat {\bf{i}} + \widehat {\bf{j}} + \left( {1 - x} \right){\bf{\hat k}}$ and ${\bf{\vec c}} = y\widehat {\bf{i}} + x\widehat {\bf{j}} + \left( {1 + x - y} \right){\bf{\hat k}}.$ Then $\left[ {{\bf{\vec a}},\;{\bf{\vec b}},\;{\bf{\vec c}}} \right]$ depends on

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Analyzing +4 | -1

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

View answer and explanation

Correct option: A

neither

x

nor

y

The correct answer is A (neither

x

nor

y

) because the scalar triple product simplifies to 1, which is independent of both

x

and

y

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},

\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}

\mathbf{b}=-\hat{\mathbf{i}}+4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}

\mathbf{c}=2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}

\mathbf{a}

a

[\mathbf{a} \mathbf{b} \mathbf{c}]