Let ${\rm{\vec A}} = {\bf{\hat i}} + {\bf{\hat j}} + {\bf{\hat k}},{\rm{\;}}{\bf{\vec B}} = {\bf{\hat i}},{\rm{\;}}{\bf{\vec C}} = {c_1}{\bf{\hat i}} + {c_2}{\bf{\hat j}} + {c_3}{\bf{\hat k}}$. If ${c_2} = - 1$ and ${c_3} = 1$, then to make three vectors coplanar

JEE MATHEMATICS MCQ: ADDITION OF VECTORS

1 Easy

1 of 1

Application +4 | -1

Let

\mathrm{\vec A} = \mathbf{\hat i} + \mathbf{\hat j} + \mathbf{\hat k},\mathrm{\;}\mathbf{\vec B} = \mathbf{\hat i},\mathrm{\;}\mathbf{\vec C} = {c_1}\mathbf{\hat i} + {c_2}\mathbf{\hat j} + {c_3}\mathbf{\hat k}

. If

{c_2} = - 1

and

{c_3} = 1

, then to make three vectors coplanar

View answer and explanation

Correct option: D

No value of

{c_1}

can be found

Option D (No value of

{c_1}

can be found) is correct because the condition for coplanarity leads to a contradiction given the values of

{c_2}

and

{c_3}

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 =