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Mathematics · 2025 · 2 marks
CBSE 2025 · Region 1 · Set 1 · Q25
(a) Two friends while flying kites from different locations, find the strings of their kites crossing each other. The strings can be represented by…
Two friends while flying kites from different locations, find the strings of their kites crossing each other. The strings can be represented by vectors $\displaystyle \vec{a}=3 \hat{i}+\hat{j}+2 \hat{k}$ and $\displaystyle \vec{b}=2 \hat{i}-2 \hat{j}+4 \hat{k}$. Determine the angle formed between the kite strings. Assume there is no slack in the strings.
Find a vector of magnitude $\displaystyle 21$ units in the direction opposite to that of $\displaystyle \overrightarrow{\mathrm{AB}}$ where A and B are the points $\displaystyle \mathrm{A}(2,1,3)$ and $\displaystyle \mathrm{B}(8,-1,0)$ respectively.
Marking-scheme solution
Let the required angle between the kite strings be $\displaystyle \boldsymbol{\theta}$.
\[\begin{aligned}
& \text { Then, } \cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \\
& \Rightarrow \cos \theta==\frac{(3 \hat{i}+\hat{j}+2 \hat{k})(2 \hat{i}-2 \hat{j}+4 \hat{k})}{\sqrt{9+1+4} \sqrt{4+4+16}}=\frac{12}{\sqrt{336}}=\frac{3}{\sqrt{21}} \\
& \Rightarrow \theta=\cos ^{-1}\left(\frac{12}{\sqrt{336}}\right) \operatorname{orcos}^{-1}\left(\frac{3}{\sqrt{21}}\right)
\end{aligned}
\]
\[\begin{aligned}
& \overrightarrow{\mathrm{B} \mathrm{A}}=-6 \hat{i}+2 \hat{j}+3 \hat{k} \\
& \text { Requiredunit vector of magnitude } 21 \\
& =21 \times\left(\frac{-6 \hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{36+4+9}}\right) \\
& =3(-6 \hat{i}+2 \hat{j}+3 \hat{k}) \text { or }-18 \hat{i}+6 \hat{j}+9 \hat{k}
\end{aligned}
\]
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