If the position vector of the vertices, $A,B,C$ of a $\Delta ABC$ are $7{\bf{\hat j}} + 10{\bf{\hat k}},{\rm{\;}} - {\bf{\hat i}} + 6{\bf{\hat j}} + 6{\bf{\hat k}}$ and $ - 4{\bf{\hat i}} + 9{\bf{\hat j}} + 6{\bf{\hat k}}$ respectively, then triangle is

JEE MATHEMATICS MCQ: ADDITION OF VECTORS

1 Easy

1 of 1

Application +4 | -1

If the position vector of the vertices,

A,B,C

of a

\Delta ABC

are

7\mathbf{\hat j} + 10\mathbf{\hat k},\mathrm{\;} - \mathbf{\hat i} + 6\mathbf{\hat j} + 6\mathbf{\hat k}

and

- 4\mathbf{\hat i} + 9\mathbf{\hat j} + 6\mathbf{\hat k}

respectively, then triangle is

View answer and explanation

Correct option: D

Right angled and isosceles also

The correct answer is D. The triangle is right angled and isosceles because two sides have equal length and satisfy the Pythagorean theorem.

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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 =