Given that vectors vector(a, b, c) form a triangle such that vector(a) = vector(b + c).

Question

Given that vectors \(\vec{a},\vec{b},\vec{c}\) form a triangle such that \(\vec{a}=\vec{b}+\vec{c}.\) Find p, q, r, s such that area of triangle is \(5\sqrt{6}\) where \(\vec{a}=p\hat{i}+q\hat{j}+r\hat{k},\) \(\vec{b}=s\hat{i}+3\hat{j}+4\hat{k}\) and \(\vec{c}=3\hat{i}+\hat{j}-2\hat{k}.\)

Answer 1

  

Given, \(\vec{a}=\vec{b}+\vec{c}\)

Equating the co-efficient of \(\hat{i},\hat{j},\hat{k}\) from both sides, we get

\(\Rightarrow s+3=p\)

\(\Rightarrow q=4\) and \(r = 2\,.....(i)\)

Now, area of triangle \(=\frac{1}{2}\bigg|\vec{b}\times \vec{c} \bigg|\)

Squaring both sides

From (i) and (ii)

\(s = -11, 5;\) \(P = -8, 8\)

\(q = 4\) and \(r = 2\)