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Mathematics · 2026 · 3 marks
CBSE 2026 · Region 5 · Set 1 · Q30
(a) Let three toys A, B and C be placed in the same straight line. If the position vectors of A, B and C are 55 i -2 j, 5 i +8 j and a i -52 j…
Let three toys A, B and C be placed in the same straight line. If the position vectors of A, B and C are $\displaystyle 55 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}, 5 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}$ and $\displaystyle \mathrm{a} \hat{\mathrm{i}}-52 \hat{\mathrm{j}}$ respectively, find the value of 'a'.
If $\displaystyle \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\displaystyle \overrightarrow{\mathrm{c}}$ are unit vectors, then prove that \[|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}|^{2}+|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|^{2} \leq 9 . \]
Marking-scheme solution
$\displaystyle \overrightarrow{AB}=-50 \hat{i}+10 \hat{j}$ and $\displaystyle \overrightarrow{BC}=(a-5) \hat{i}-60 \hat{j}$
As $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{BC}$ are collinear vectors
$\displaystyle \therefore \dfrac{-50}{a-5}=\dfrac{10}{-60}$
$\displaystyle a=305$
$\displaystyle |\vec{a}|=|\vec{b}|=|\vec{c}|=1$.
Now, consider
$\displaystyle |\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}$
$\displaystyle =2|\vec{a}|^{2}+2|\vec{b}|^{2}+2|\vec{c}|^{2}-2 \vec{a} \cdot \vec{b}-2 \vec{b} \cdot \vec{c}-2 \vec{c} \cdot \vec{a}$
$\displaystyle =3|\vec{a}|^{2}+3|\vec{b}|^{2}+3|\vec{c}|^{2}-\left(|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot \vec{c}+2 \vec{c} \cdot \vec{a}\right)$
$\displaystyle =9-|\vec{a}+\vec{b}+\vec{c}|^{2} \leq 9$
$\displaystyle \therefore|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2} \leq 9$
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