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Mathematics · 2024 · 4 marks
CBSE 2024 · Region 4 · Set 1 · Q37
An instructor at the astronomical centre shows three among the brightest stars in a particular constellation. Assume that the telescope is located at…
An instructor at the astronomical centre shows three among the brightest stars in a particular constellation. Assume that the telescope is located at $\displaystyle \mathrm{O}(0,0,0)$ and the three stars have their locations at the points $\displaystyle \mathrm{D}, \mathrm{A}$ and V having position vectors $\displaystyle 2 \hat{i}+3 \hat{j}+4 \hat{k}, 7 \hat{i}+5 \hat{j}+8 \hat{k}$ and $\displaystyle -3 \hat{i}+7 \hat{j}+11 \hat{k}$ respectively.
Based on the above information, answer the following questions :
(i)
How far is the star V from star A ?
(ii)
Find a unit vector in the direction of $\displaystyle \overrightarrow{\mathrm{DA}}$.
(iii)
Find the measure of $\displaystyle \angle \mathrm{VDA}$.
OR
(iii)
What is the projection of vector $\displaystyle \overrightarrow{\mathrm{DV}}$ on vector $\displaystyle \overrightarrow{\mathrm{DA}}$ ?
Marking-scheme solution
(i)
$\displaystyle \overrightarrow{\mathrm{A} V}=$ Position Vector of V - Position Vector of A
$$=-$\displaystyle 3$ \hat{i}+$\displaystyle 7$ \hat{j}+$\displaystyle 11$ \hat{k}-$\displaystyle 7$ \hat{i}-$\displaystyle 5$ \hat{j}-$\displaystyle 8$ \hat{k}=-$\displaystyle 10$ \hat{i}+$\displaystyle 2$ \hat{j}+$\displaystyle 3$ \hat{k}
$$Thus, $\displaystyle |\overrightarrow{\mathrm{A} V}|=\sqrt{100+4+9}=\sqrt{113}$ units
(ii)
$\displaystyle \overrightarrow{\mathrm{D} \mathrm{A}}=$ Position Vector of $\displaystyle \mathrm{A}-$ Position Vector of D
$$=$\displaystyle 7$ \hat{i}+$\displaystyle 5$ \hat{j}+$\displaystyle 8$ \hat{k}-$\displaystyle 2$ \hat{i}-$\displaystyle 3$ \hat{j}-$\displaystyle 4$ \hat{k}=$\displaystyle 5$ \hat{i}+$\displaystyle 2$ \hat{j}+$\displaystyle 4$ \hat{k}
$$Unit vector in the direction of $\displaystyle \overrightarrow{\mathrm{D} \mathrm{A}}=\frac{5 \hat{i}+2 \hat{j}+4 \hat{k}}{3 \sqrt{5}}$
$\displaystyle \overrightarrow{\mathrm{D} V}=-5 \hat{i}+4 \hat{j}+7 \hat{k}$
$\displaystyle \angle V \mathrm{D} \mathrm{A}=\cos ^{-1}\left(\frac{\overrightarrow{\mathrm{D} V} \cdot \overrightarrow{\mathrm{D} \mathrm{A}}}{|\overrightarrow{\mathrm{D} V}||\overrightarrow{\mathrm{D} \mathrm{A}}|}\right)=\cos ^{-1}\left(\frac{11 \sqrt{2}}{90}\right)$
$\displaystyle \overrightarrow{\mathrm{D} V}=-5 \hat{i}+4 \hat{j}+7 \hat{k}$
Projection of $\displaystyle \overrightarrow{\mathrm{D} V}$ on $\displaystyle \overrightarrow{\mathrm{D} \mathrm{A}}=\left(\frac{\overrightarrow{\mathrm{D} V} \cdot \overrightarrow{\mathrm{D} \mathrm{A}}}{|\overrightarrow{\mathrm{D} \mathrm{A}}|}\right)=\frac{11 \sqrt{5}}{15}$
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