Show that the points 2 i ,- i -4 j and - i +4 j form an isosceles triangle.

Given:- The points A, B, C with position vectors a , b , c respectively. Also, a =2 i b =- i -4 j c =- i +4 j Then, AB = b - a AB = (- i -4 j )-2 i AB =-3 i -4 j Now, | AB |=(-3)^2+(-4)^2 =9+16 =25 =5 BC = c - b BC = (- i +4 j )- (- i -4 j ) BC =- i +4 j + i +4 j BC =8 j

Show that the points 2 i ,- i -4 j and - i +4 j form an isosceles…

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Question

Show that the points $2\hat{\text{i}},-\hat{\text{i}}-4\hat{\text{j}}\text{ and }-\hat{\text{i}}+4\hat{\text{j}}$ form an isosceles triangle.

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Answer

Given:- The points A, B, C with position vectors $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ respectively.
Also, $\vec{\text{a}}=2\hat{\text{i}}$
$\vec{\text{b}}=-\hat{\text{i}}-4\hat{\text{j}}$
$\vec{\text{c}}=-\hat{\text{i}}+4\hat{\text{j}}$
Then,
$\overrightarrow{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}$
$\Rightarrow\overrightarrow{\text{AB}}=\big(-\hat{\text{i}}-4\hat{\text{j}}\big)-2\hat{\text{i}}$
$\Rightarrow\overrightarrow{\text{AB}}=-3\hat{\text{i}}-4\hat{\text{j}}$
Now, $\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{(-3)^2+(-4)^2}$
$=\sqrt{9+16}$
$=\sqrt{25}$
$=5$
$\overrightarrow{\text{BC}}=\vec{\text{c}}-\vec{\text{b}}$
$\Rightarrow\overrightarrow{\text{BC}}=\big(-\hat{\text{i}}+4\hat{\text{j}}\big)-\big(-\hat{\text{i}}-4\hat{\text{j}}\big)$
$\Rightarrow\overrightarrow{\text{BC}}=-\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{i}}+4\hat{\text{j}}$
$\Rightarrow\overrightarrow{\text{BC}}=8\hat{\text{j}}$

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