Prove that the points i - j , 4 i +3 j + k and 2 i -4 j +5 k are the vertices of a right-angled triangle.

Let A = i - j B =4 i +3 j + k C =2 i -4 j +5 k AB = B - A = (4 i +3 j + k )- ( i - j ) =4 i +3 j + k - i - j =3 i +4 j + k | AB |=(3^2)+(4)^2+(1)^2 =9+16+1 =26 BC = C - B = ( 2i -4 j +5 k )- (4 i +3 j + k ) =2 i -4 j +5 k -4 i -3 j - k =-2 i -7 j +4 k | BC |=(2)^2+(-7)^2+(4)^2 =4+49+16 =69 CA = A - C = i - j - (2 i -4 j +5 k ) = i - j -2 i +4 j -5 k =- i +3 j -5 k | CA |=(-1)^2+(3)^2+(-5)^2 =1+9+25 =35 Here, | AB |^2+ | CA |^2= | BC |^2 26+35=69 61 69 LHS Since sum of square of two sides is not equal to the square of third sides. So, is not a right triangle.

Prove that the points i - j , 4 i +3 j + k and 2 i -4 j +5 k are …

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Question

Prove that the points $\hat{\text{i}}-\hat{\text{j}},\ 4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$ and $2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$ are the vertices of a right-angled triangle.

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Answer

Let $\vec{\text{A}}=\hat{\text{i}}-\hat{\text{j}}$
$\vec{\text{B}}=4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{C}}=2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$
$\overrightarrow{\text{AB}}=\vec{\text{B}}-\vec{\text{A}}$
$=\big(4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\big)-\big(\hat{\text{i}}-\hat{\text{j}}\big)$
$=4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}-\hat{\text{i}}-\hat{\text{j}}$
$=3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}}$
$\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{(3^2)+(4)^2+(1)^2}$
$=\sqrt{9+16+1}$
$=\sqrt{26}$
$\overrightarrow{\text{BC}}=\vec{\text{C}}-\vec{\text{B}}$
$=\big(\hat{2\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\big)-\big(4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\big)$
$=2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}-4\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}}$
$=-2\hat{\text{i}}-7\hat{\text{j}}+4\hat{\text{k}}$
$\Big|\overrightarrow{\text{BC}}\Big|=\sqrt{(2)^2+(-7)^2+(4)^2}$
$=\sqrt{4+49+16}$
$=\sqrt{69}$
$\overrightarrow{\text{CA}}=\vec{\text{A}}-\vec{\text{C}}$
$=\hat{\text{i}}-\hat{\text{j}}-\big(2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\big)$
$=\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
$=-\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}$
$\Big|\overrightarrow{\text{CA}}\Big|=\sqrt{(-1)^2+(3)^2+(-5)^2}$
$=\sqrt{1+9+25}$
$=\sqrt{35}$
Here, $\Big|\overrightarrow{\text{AB}}\Big|^2+\Big|\overrightarrow{\text{CA}}\Big|^2=\Big|\overrightarrow{\text{BC}}\Big|^2$
$26+35=69$
$61\neq69$
$\text{LHS}\neq\text{RHS}$
Since sum of square of two sides is not equal to the square of third sides. So, $\triangle\text{ABC}$ is not a right triangle.

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