JEE MATHEMATICS MCQ: ADDITION OF VECTORS

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Analyzing +4 | -1

Let

\vec{a}, \vec{b}

and

\vec{c}

be three non-zero vectors such that no two of these are collinear. If the vector

\vec{a}+2 \vec{b}

is collinear with

\vec{c}

and

\vec{b}+3 \vec{c}

is collinear with

\vec{a}

(

\lambda

being some non-zero scalar), then

\vec{a}+2 \vec{b}+6 \vec{c}

equals

View answer and explanation

Correct option: D

\overrightarrow{0}

Option D (

\overrightarrow{0}

) is correct because by manipulating the collinearity conditions, we find that

\vec{a}+2 \vec{b}+6 \vec{c}

equals the zero vector.

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 =