Show that the points A 2- - - - k- B - - 3- - 5k- and C 3- - 4- - 4k- are the vertices of a right angled triangle

Given #10;that the points let O be the position vector then, #10; #10; OA( #10;2i j+k),OB( #10;i 3j 5k) #160; #160;and #160; #160; ...

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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

Show that the...

Question

Show that the points $A (2^i -^i +^k) B (^i - 3^j - 5^k)$ and $C (3^i - 4^j - 4^k)$ are the vertices of a right angled triangle

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Solution

Given that the points let $O$ be the position vector then,

$O A (2 ˆ i - ˆ j + ˆ k), O B (ˆ i - ˆ 3 j - 5 ˆ k)$ and $O C (3 ˆ i - ˆ 4 j - 4 ˆ k)$

Prove that all given points are the vertices of aright angle triangle.

Proof:-

$\to A . \to B = \to O \to B - - \to - O \to A$

$= (ˆ i - 3 ˆ j - 5 ˆ k) - (2 ˆ i - ˆ j + ˆ k)$

$= - ˆ i - 2 ˆ j - 6 ˆ k$

Now,

$\to B . \to C = \to O \to C - \to O \to B$

$= (3 ˆ i - 4 ˆ j - 4 ˆ k) - (ˆ i - ˆ 3 j - 5 ˆ k)$

$= 2 ˆ i - ˆ j + ˆ k$

And

$\to C . \to A = \to O \to A - \to O \to C$

$= (2 ˆ i - ˆ j + ˆ k) - (3 ˆ i - 4 ˆ j - 4 ˆ k)$

$= - ˆ i + 3 ˆ j + 5 ˆ k$

We know that the two vectors are perpendicular to each other

If their scalar product is zero.

Then,

$\to B \to C . \to C \to A = (2 ˆ i - ˆ j + ˆ k) . (- ˆ i + 3 ˆ j + 5 ˆ k)$

$= (2 \times - 1) + (- 1 \times 3) + (1 \times 5)$

$= - 2 - 3 + 5$

$= - 5 + 5$

$= 0$

Since, $[\to B \to C . \to C \to A] = 0$

Then $[\to B \to C]$ is perpendicular to $[\to C \to A]$

Hence given $Δ A B C$ is right angle.

Hence proved.

Suggest Corrections

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Show that the points A(2^i−^i+^k)B(^i−3^j−5^k) and C(3^i−4^j−4^k) are the vertices of a right angled triangle

Show that the points $A (2^i -^i +^k) B (^i - 3^j - 5^k)$ and $C (3^i - 4^j - 4^k)$ are the vertices of a right angled triangle