Let $A, B, C$ be three points whose position vectors respectively are $\vec{a}=\hat{i}+4 \hat{j}+3 \hat{k}$ $\vec{b}=2 \hat{i}+\alpha \hat{j}+4 \hat{k}, \alpha \in \mathbb{R}$ $\vec{c}=3 \hat{i}-2 \hat{j}+5 \hat{k}$ If $\alpha$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non collinear, then the length of the median, in $\triangle A B C$, through $A$ is:

JEE MATHEMATICS MCQ: ADDITION OF VECTORS

1 Easy

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Analyzing +4 | -1

Let

A, B, C

be three points whose position vectors respectively are

\vec{a}=\hat{i}+4 \hat{j}+3 \hat{k}

\vec{b}=2 \hat{i}+\alpha \hat{j}+4 \hat{k}, \alpha \in \mathbb{R}

\vec{c}=3 \hat{i}-2 \hat{j}+5 \hat{k}

\alpha

is the smallest positive integer for which

\vec{a}, \vec{b}, \vec{c}

are non collinear, then the length of the median, in

\triangle A B C

, through

A

is:

View answer and explanation

Correct option: A

\frac{\sqrt{82}}{2}

Option A (

\frac{\sqrt{82}}{2}

) is correct because it is the calculated length of the median AM after determining the smallest positive integer value of

\alpha

for which the points are non-collinear.

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 =