Question

If the position vectors of the vertices of a triangle be 6i+4j+5k, 4i+5j+6k and 5i+6j+4k then the triangle is
(A) Right angled triangle
(B) Equilateral triangle
(C) Isosceles triangles
(D) None of these

Answer 1

Hint-In this question we are given that we have to find the type of triangle given that position vectors of the vertices are 6i+4j+5k, 4i+5j+6k and 5i+6j+4k.In this type of question we have to find the length of the sides by its position vectors so we look for each and every separately by dividing it on basis of its vectors and we can easily find them.

Complete step-by-by-solution -

Let the vertices of triangle is
$\overrightarrow {OA} = (\overrightarrow {6i} + \overrightarrow {4j} + \overrightarrow {5k} ) \\\ \overrightarrow {OB} = (\overrightarrow {4i} + \overrightarrow {5j} + \overrightarrow {6k} ) \\\ \overrightarrow {OC} = (\overrightarrow {5i} + \overrightarrow {6j} + \overrightarrow {4k} ) \\\ Now, \\\ \overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} \\\$

$\overrightarrow {AB} = (\overrightarrow {4i} + \overrightarrow {5j} + \overrightarrow {6k} ) - (\overrightarrow {6i} + \overrightarrow {4j} + \overrightarrow {5k} ) \\\ \overrightarrow {AB} = \overrightarrow {4i} + \overrightarrow {5j} + \overrightarrow {6k} - \overrightarrow {6i} - \overrightarrow {4j} - \overrightarrow {5k} \\\ \overrightarrow {AB} = - \overrightarrow {2i} + \overrightarrow j + \overrightarrow k \\\ \overrightarrow {BC} = \overrightarrow {OC} - \overrightarrow {OB} \\\ \overrightarrow {BC} = (\overrightarrow {5i} + \overrightarrow {6j} + \overrightarrow {4k} ) - (\overrightarrow {4i} + \overrightarrow {5j} + \overrightarrow {6k} ) \\\$

$\overrightarrow {BC} = \overrightarrow {5i} + \overrightarrow {6j} + \overrightarrow {4k} - \overrightarrow {4i} - \overrightarrow {5j} - \overrightarrow {6k} \\\ \overrightarrow {BC} = \overrightarrow i + \overrightarrow j - \overrightarrow {2k} \\\ \overrightarrow {CA} = \overrightarrow {OA} - \overrightarrow {OC} \\\ \overrightarrow {CA} = (\overrightarrow {6i} + \overrightarrow {4j} + \overrightarrow {5k} ) - (\overrightarrow {5i} + \overrightarrow {6j} + \overrightarrow {4k} ) \\\ \overrightarrow {CA} = \overrightarrow {6i} + \overrightarrow {4j} + \overrightarrow {5k} - \overrightarrow {5i} - \overrightarrow {6k} - \overrightarrow {4k} \\\ \overrightarrow {CA} = \overrightarrow i - \overrightarrow {2j} + \overrightarrow k \\\$
Now,
$\left| {AB} \right| = \sqrt {{{( - 2)}^2} + {1^2} + {1^2}} = \sqrt 6 \\\ \left| {BC} \right| = \sqrt {{1^2} + {1^2} + {{( - 2)}^2}} = \sqrt 6 \\\ \left| {CA} \right| = \sqrt {{1^2} + {{( - 2)}^2} + {1^2}} = \sqrt 6 \\\$
Then,$AB = BC = CA = \sqrt 6$
Hence, it is an equilateral triangle.
Thus option B is correct.

Note-In this question it is to be noted that we have to answer the question by finding its sides provided that the position vectors of the vertices of a triangle be 6i+4j+5k, 4i+5j+6k and 5i+6j+4k
So first as you can see I found for each side differently by dividing it by vector O. Thus we were able to find the sides and able to determine it is an equilateral triangle.