If the vectors ${\bf{\vec a}} = 2\widehat {\bf{i}} + 3\widehat {\bf{j}} + 6{\bf{\hat k}}$ and ${{\bf{\vec b}}}$ are collinear and $\left| {{\bf{\vec b}}} \right| = 21,\;{\rm{then\;}}{\bf{\vec b}}$ is equal to

$ \pm 3\left( {2\widehat {\bf{i}} + 3\widehat {\bf{j}} + 6{\bf{\hat k}}} \right)$

$ \pm \left( {2\widehat {\bf{i}} + 3\widehat {\bf{j}} + 6{\bf{\hat k}}} \right)$

$\left( {\widehat {\bf{i}} + \widehat {\bf{j}} + {\bf{\hat k}}} \right)$

$ \pm 21\left( {2\widehat {\bf{i}} + 3\widehat {\bf{j}} + 6{\bf{\hat k}}} \right)$

JEE MATHEMATICS MCQ: ADDITION OF VECTORS

1 Easy

1 of 1

Application +4 | -1

If the vectors

\mathbf{\vec a} = 2\widehat \mathbf{i} + 3\widehat \mathbf{j} + 6\mathbf{\hat k}

and

{\mathbf{\vec b}}

are collinear and

\left| {\mathbf{\vec b}} \right| = 21,\;\mathrm{then\;}\mathbf{\vec b}

is equal to

View answer and explanation

Correct option: B

\pm 3\left( {2\widehat \mathbf{i} + 3\widehat \mathbf{j} + 6\mathbf{\hat k}} \right)

Option B (

\pm 3\left( {2\widehat \mathbf{i} + 3\widehat \mathbf{j} + 6\mathbf{\hat k}} \right)

) is correct because collinear vectors are scalar multiples of each other, and the magnitude condition determines the scalar.

Practice All Chapter MCQs →

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

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

\mathbf{d}

a

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

\mathbf{d}

{\mathbf{\vec a}}

2\;p

{\mathbf{\vec a}}

p + 1

\vec{a}, \vec{b}

\vec{c}

\vec{d}

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

\overrightarrow{\mathrm{c}}

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

\vec{a}, \vec{b}

\vec{c}

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

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

\vec{a}

\vec{c}

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

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

\lambda

AD

ABC

E,\;BE

AC

F

AF:AC =