CBSE 2022 · Region 4 · Set 1 · Q4 · 2 marks
If $\displaystyle |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|^{2}=400$ and $\displaystyle |\overrightarrow{\mathrm{b}}|=5$, then find the value of $\displaystyle |\overrightarrow{\mathrm{a}}|$.Find all the possible vectors of magnitude $\displaystyle 5 \sqrt{3}$ which are equally inclined to the coordinate axes.
If $\displaystyle |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|^{2}=400$ and $\displaystyle |\overrightarrow{\mathrm{b}}|=5$, then find the value of $\displaystyle |\overrightarrow{\mathrm{a}}|$.
Find all the possible vectors of magnitude $\displaystyle 5 \sqrt{3}$ which are equally inclined to the coordinate axes.
Marking-scheme solution
(a) \(\displaystyle |\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta=400\)
\(\displaystyle |\vec{a}|^{2} \cdot 25(1)=400\)
\(\displaystyle |\vec{a}|^{2}=16\)
\(\displaystyle |\vec{a}|=4\)
Or(b) Let the required vector be \(\displaystyle x \hat{i}+x \hat{j}+x \hat{k}\)
\(\displaystyle \sqrt{3 x^{2}}=5 \sqrt{3}\)
\(\displaystyle x^{2}=25 \Rightarrow x= \pm 5\)
Required vectors are \(\displaystyle 5 \hat{i}+5 \hat{j}+5 \hat{k}\) or \(\displaystyle -5 \hat{i}-5 \hat{j}-5 \hat{k}\).Vector AlgebraProduct of Two VectorsApplyvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.