Let $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{c}=\hat{\mathbf{k}}-\hat{\mathbf{i}}$. If $\mathbf{d}$ is $a$ unit vector such that $\mathbf{a} \cdot \mathbf{d}=0=(b \mathbf{c}$ ) , then $\mathbf{d}$ is equal to

$\pm \frac{(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})}{\sqrt{3}}$

$\pm \frac{(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{3}}$

JEE MATHEMATICS MCQ: ADDITION OF VECTORS

1 Medium

1 of 1

Application +4 | -1

Let

\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}-\hat{\mathbf{k}}

and

\mathbf{c}=\hat{\mathbf{k}}-\hat{\mathbf{i}}

. If

\mathbf{d}

a

unit vector such that

\mathbf{a} \cdot \mathbf{d}=0=(b \mathbf{c}

) , then

\mathbf{d}

is equal to

View answer and explanation

Correct option: A

\pm \frac{(\hat{i}+\hat{j}-2 \hat{k})}{\sqrt{6}}

Option A is correct because the unit vector

\mathbf{d}

is perpendicular to

\mathbf{a}

and lies in the plane of

\mathbf{b}

and

\mathbf{c}

, which leads to

\mathbf{d} = \pm \frac{(\hat{i}+\hat{j}-2 \hat{k})}{\sqrt{6}}

More Questions from ADDITION OF VECTORS

Practice All Chapter MCQs →

A vector

{\mathbf{\vec a}}

has components

2\;p

and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense, if this respect to new system

{\mathbf{\vec a}}

has components

p + 1

and 1, then If

\vec{a}, \vec{b}

and

\vec{c}

are three mutually perpendicular unit vetors and

\vec{d}

is a unit vector which makes equal angles with

\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}

ānd

\overrightarrow{\mathrm{c}}

then the value of

|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|^2

The non-zero vectors are

\vec{a}, \vec{b}

and

\vec{c}

are related by

\vec{a}=8 \vec{b}

and

\vec{c}=-7 \vec{b}

. Then the angle between

\vec{a}

and

\vec{c}

is If the vectors

2 \mathbf{i}+\mathbf{j}-\mathbf{k},-\mathbf{i}+2 \mathbf{j}+\lambda \mathbf{k}

and

-5 \mathbf{i}+2 \mathbf{j}-\mathbf{k}

ar coplanar, then the value of

\lambda

is equal The medium

AD

of the triangle

ABC

is bisected at

E,\;BE

meets

AC

F

, then

AF:AC =

If a vector

{\vec r}

of magnitude

3\sqrt 6

is directed along the bisector of the angle between the vectors

\vec a = 7\hat i - 4\hat j - 4\hat k

and

\vec b = - 2\hat i - \hat j + 2\hat k

, then

\vec r =