JEE MATHEMATICS MCQ: VECTOR TRIPLE PRODUCT

1 Easy

1 of 1

Application +4 | -1

Let

P, Q, R

be points with position vectors

{r}_1=3 {i}-2 {j}-{k}, {r}_2={i}+3 {j}+4{k}

and

{r}_3=2 {i}+{j}-2 {k}

relative to an origin 0. The distance of

P

from the plane

O Q R

is (magnitude)

View answer and explanation

Correct option: B

Option B (3) is correct because the distance of point P from the plane OQR is calculated using the formula involving the scalar triple product and equals 3.

More Questions from VECTOR TRIPLE PRODUCT

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Let

{\mathbf{\hat u}}

and

{\mathbf{\hat v}}

are unit vectors such that

\mathbf{\hat u}\cdot\mathbf{\hat v} = 0

{\mathbf{\hat r}}

is any vector coplanar with

{\mathbf{\hat u}}

and

{\mathbf{\hat v}}

, then the magnitude of the vector

\mathbf{\vec r} \times \left( {\mathbf{\hat u} \times \mathbf{\hat v}} \right)

is Let

\lambda \in \mathrm{R}, \vec{a}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}, \vec{b}=\hat{i}-\lambda \hat{j}+2 \hat{k}

((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}

then

|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2

is equal to Let

\vec{a}=-\hat{i}-\hat{j}+\hat{k}

such that

\vec{a} \cdot \vec{b}=1

and

\vec{a} \times \vec{b}=\hat{i}-\hat{j}

then

\vec{a}-6 \vec{b}

equals Let

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

be any three vectors. Then, vectors

\vec u = \vec a \times (\vec b \times \vec c),\vec v = \vec b \times (\vec c \times \vec a)\,\,\,{\text{and}}\,\,\,\vec w = \vec c \times (\vec a \times \vec b)

are such that they are If

a

and

b

are perpendicular, then

a\, \times (a\, \times (a\, \times (a\, \times b)))

is equal to For any vector

\vec{a}

, the value of

\hat{i} \times(\vec{a} \times \vec{i})+\hat{j} \times(\vec{a} \times \hat{j})+\hat{k} \times(\vec{a} \times \vec{k})

, is