JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Hard

1 of 1

Application +4 | -1

A vector

\mathbf{n}

of magnitude 8 units is inclined to

x

-axis at

45^{\circ}, y

-axis at

60^{\circ}

and an acute angle with

z

-axis. If a plane passes through a point

(\sqrt{2},-1,1)

and is normal to

\mathbf{n}

, then its equation in vector form is

View answer and explanation

Correct option: B

\mathbf{r} \cdot(\sqrt{2} \mathbf{i}+\mathbf{j}+\mathbf{k})=2

Option B (

\mathbf{r} \cdot(\sqrt{2} \mathbf{i}+\mathbf{j}+\mathbf{k})=2

) is correct because it is the equation of the plane passing through the given point and normal to the vector

\mathbf{n}

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}