$\vec{a}$ and $\vec{c}$ are unit vectors and $|\vec{b}|=4$. If angle between $\vec{b}$ and $\vec{c}$ is $\cos ^{-1}\left(\frac{1}{4}\right)$ and $\vec{a} \times \vec{b}=2 \vec{a} \times \vec{c}$, then $\vec{b}=\lambda \vec{a}+2 \vec{c}$, where $\lambda$ is equal to

JEE MATHEMATICS MCQ: ADDITION OF VECTORS

1 Medium

1 of 1

Application +4 | -1

\vec{a}

and

\vec{c}

are unit vectors and

|\vec{b}|=4

. If angle between

\vec{b}

and

\vec{c}

\cos ^{-1}\left(\frac{1}{4}\right)

and

\vec{a} \times \vec{b}=2 \vec{a} \times \vec{c}

, then

\vec{b}=\lambda \vec{a}+2 \vec{c}

, where

\lambda

is equal to

View answer and explanation

Correct option: D

\pm 4

Option D (

\pm 4

) is correct because by using the given conditions and vector properties, we can derive that

\lambda = \pm 4

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 =