JEE MATHEMATICS MCQ: ADDITION OF VECTORS

1 Medium

1 of 1

Application +4 | -1

\bar{a}=-\bar{i}+\bar{j}, \bar{b}=\bar{j}-\bar{k}, \bar{c}=\bar{i}+\bar{j}+\bar{k}

then the unit vector in the opposite direction of

\bar{a}+2 \bar{b}-3 \bar{c}

View answer and explanation

Correct option: C

\frac{4 \bar{i}+5 \bar{k}}{\sqrt{41}}

Option C (

\frac{{4 \bar{{i}}+5 \bar{{k}}}}{{\sqrt{{41}}}}

) is correct because it represents the unit vector in the opposite direction of the calculated vector

\bar{{a}}+2 \bar{{b}}-3 \bar{{c}}

More Questions from ADDITION OF VECTORS

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 =