Question

Two friends while flying kites from different locations, find the strings of their kites crossing each other. The strings can be represented by vectors \(\vec{a} = 3\hat{i}+\hat{j} +2\hat{k}\) and \(\vec{b} = 2\hat{i} - 2\hat{j}+4\hat{k}\) Determine the angle formed between the kite strings. Assume there is no slack in the strings.

OR

(b) Find a vector of magnitude 21 units in the direction opposite to that of \(\vec{AB}\) where A and B are the points A(2, 1, 3) and B(8, –1, 0) respectively.

Answer 1

Since there is no slack in the string, strings can be taken as straight lines with direction as given vectors. The angles between two vectors v₁ and v₂ is given by \(\frac{\vec{V_1} \cdot \vec{V_2}}{|\vec{V_1}||\vec{V_2}|} = cos \theta\)

⇒ cosine of Angle between \(\vec{a}\) and \(\vec{b}\)will be \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\) 

\(|\vec{a}| = \sqrt{{3}^2 +{1}^2+{2}^2} = \sqrt{9+1+4} = \sqrt14\) 

\(|\vec{b}| = \sqrt{{2}^2+(-2)^2+{4}^2} = \sqrt{4+4+16} =\sqrt24\) 

\(\vec{a} \cdot \vec{b} = (3\hat{i}+\hat{j}+2\hat{k}) \cdot (2\hat{i}-2\hat{j}+4\hat{k}) = 2 \times 3+(1)(-2)+(2)(4)\) 

= 6 - 2 + 8 = 12

\(cos\theta = \frac{12}{\sqrt{14}\sqrt{24}} = \sqrt{\frac{144}{24 \times 14}} = \sqrt{\frac{3}{7}}\) 

⇒ \(\theta = cos^{-1}(\sqrt{\frac{3}{7}})\)