CBSE 2025 · Region 6 · Set 1 · Q24 · 2 marks
If $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ are position vectors of point $\displaystyle A$ and point $\displaystyle B$ respectively, find the position vector of point C on BA produced such that $\displaystyle \mathrm{BC}=3 \mathrm{BA}$.Vector $\displaystyle \vec{r}$ is inclined at equal angles to the three axes $\displaystyle x$, $\displaystyle y$ and $\displaystyle z$. If magnitude of $\displaystyle \vec{r}$ is $\displaystyle 5 \sqrt{3}$ units, then find $\displaystyle \vec{r}$.
If $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ are position vectors of point $\displaystyle A$ and point $\displaystyle B$ respectively, find the position vector of point C on BA produced such that $\displaystyle \mathrm{BC}=3 \mathrm{BA}$.
Vector $\displaystyle \vec{r}$ is inclined at equal angles to the three axes $\displaystyle x$, $\displaystyle y$ and $\displaystyle z$. If magnitude of $\displaystyle \vec{r}$ is $\displaystyle 5 \sqrt{3}$ units, then find $\displaystyle \vec{r}$.
Marking-scheme solution
C divides BA in the ratio $\displaystyle 3: 2$ externally Required vector $\displaystyle =\overrightarrow{\mathrm{c}}=\frac{3 \vec{a}-2 \vec{b}}{3-2}=3 \vec{a}-2 \vec{b}$
Unit vector equally inclined along coordinate axes is $\displaystyle \frac{\hat{\mathrm{i}}}{\sqrt{3}}+\frac{\hat{\mathrm{j}}}{\sqrt{3}}+\frac{\hat{\mathrm{k}}}{\sqrt{3}} \vec{r}=5 \sqrt{3}\left(\frac{\hat{\mathrm{i}}}{\sqrt{3}}+\frac{\hat{\mathrm{j}}}{\sqrt{3}}+\frac{\hat{\mathrm{k}}}{\sqrt{3}}\right)=5 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} \quad($ or $\displaystyle -5 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})$
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.