Let $ABC$ be a triangle, the position vectors of whose vertices are respectively $7\widehat {\bf{i}} + 10{\bf{\hat k}},\; - \widehat {\bf{i}} + 6\widehat {\bf{j}} + 6{\bf{\hat k}}$ and $ - 4\widehat {\bf{i}} + 9\widehat {\bf{j}} + 6{\bf{\hat k}}$ Then, the $\Delta ABC$ is

JEE MATHEMATICS MCQ: ADDITION OF VECTORS

1 Easy

1 of 1

Application +4 | -1

Let

ABC

be a triangle, the position vectors of whose vertices are respectively

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

and

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

Then, the

\Delta ABC

View answer and explanation

Correct option: C

Right angled isosceles

Option C (Right angled isosceles) is correct because the triangle has two sides of equal length and satisfies the Pythagorean theorem.

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 =