The position vectors of points A,B and C are `hati+hatj,hati + 5hatj -hatk and 2hati + 3hatj + 5hatk`, respectively the greatest angle of triangle ABC is

Options

A. `120^(@)`
B. `90^(@)`
C. `cos^(-1)(3//4)`
D. none of these

Answer

b

Explanation

Since `vec(OA) = hati +hatj + hatk`
` vec(OB) = hati + 5hatj -hatk`
`vec(OC) = 2hati + 3hatj + 5hatk`
` a = BC |vec(BC)|= |vec(OC) - vec(OB)|`
`|hati - 2hatj + 6hatk| = sqrt41`
`b = CA= |vec(CA)|= |vec(OA) -vec(OC)|`
` = | -hati - 2hatj - 4hatk| = sqrt21`
`and c= AB= |vec(AB)| = |vec(OB)-vec(OA)|`
`|0 hati + 4hatj - 2hatk| =sqrt20`
Since `a gt b gt c`, A is the greatest angle l. therefore,
` cos A = (b^(2) + c^(2)-a^(2))/2bc) = (21 + 20 -41)/(2. sqrt21 . sqrt20) = 0`
`angleA = 90^(@)`