The vectors $2\,\hat i + 3\hat j,\;\,5\,\hat i + 6\hat j$ and $8\,\hat i + \lambda \hat j$ have their initial points at $\left( {{\rm{1, 1}}} \right)$. The value of $\lambda $ so that the vectors terminate on one straight line, is

JEE MATHEMATICS MCQ: ADDITION OF VECTORS

1 Easy

1 of 1

Application +4 | -1

The vectors

2\,\hat i + 3\hat j,\;\,5\,\hat i + 6\hat j

and

8\,\hat i + \lambda \hat j

have their initial points at

\left( {\mathrm{1, 1}} \right)

. The value of

\lambda

so that the vectors terminate on one straight line, is

View answer and explanation

Correct option: D

Option D (9) is correct because the value of

\lambda

is found by ensuring that the terminal points of the given vectors are collinear, which leads to a linear equation that can be solved for

\lambda

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\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 =