Question

show that vector a=3i-2j+k , vector b=i-3j+5k and vector c=2i+j-4k forms a right angled triangle?

Answer 1

Here a^2 +c^2 = b^2i.e.
(3i-2i+k)^2 +(2i+j-4k)^2=(i-3j+5k)^2
9+4+1+2(0+0+0) + 4+1+16+2(0+0+0)=
1+9+25
35=35 Therefore a,b,c are sides od right angled triangle.

Answer 2

Given:

Show that vector a=3i-2j+k , vector b=i-3j+5k and vector c=2i+j-4k forms a right angled triangle

Solution:

a=3i-2j+k , vector b=i-3j+5k and vector c=2i+j-4k

$A^2 + c^2 = b^2$

$(3i-2j+k)^2 + (i-3j+5k)^2 = (2i+j-4k)^2$

(9 + 4 + 1) + 4 + 1 + 16 = 1 + 9 + 25

35 = 35

Both the sides are equal.

Therefore a, b and c are the sides of the right angled triangle .

To Know more:

A = 2i - j + k, b = i - 3j - 5k. Find the vector c such that a, b and c form the sides of a triangle.

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