Question

If $OABCD$ is a pyramid with square base $ABCD$ of unit side and vertex $O$ such that $OA=OB=OC=OD=1$ unit, then

• A. angle between the line $AC$ and $OD$ is $\frac{\pi }{3}.$
• B. angle between the line $AC$ and $OD$ is $\frac{\pi }{2}.$
• C. shortest distance between $AC$ and $OD$ is $\frac{1}{\sqrt{2}}.$ unit
• D. shortest distance between $AC$ and $OD$ is $\frac{1}{2}$ unit

Answer 1

Answer: (B), (D)

shortest distance between $AC$ and $OD$ is $\frac{1}{2}$ unit

Let $A\left(\overrightarrow{0}\right),B\left(\hat{i}\right),$ and $D\left(\hat{j}\right)$
Then, $C(\hat{i}+\hat{j})$
Let $M$ be tha mid point of $AC.$
Then, $\overrightarrow{M}=\frac{\hat{i}+\hat{j}}{2}$
Since $OA=1$ and $AM=\frac{1}{\sqrt{2}}$
$\therefore OM=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$
Thus, position vector of $\overrightarrow{O}=\frac{\hat{i}+\hat{j}}{2}+\frac{\hat{k}}{\sqrt{2}}$

Now, $\overrightarrow{OD}=\hat{j}-(\frac{\hat{i}+\hat{j}}{2}+\frac{\hat{k}}{\sqrt{2}})$
$\Rightarrow \overrightarrow{OD}=\frac{-\hat{i}+\hat{j}}{2}-\frac{\hat{k}}{\sqrt{2}}$ and
$\overrightarrow{AC}=\hat{i}+\hat{j}$
Since $\overrightarrow{OD}\cdot \overrightarrow{AC}=0$
$\therefore$ Angle between $AC$ and $OD$ is $\frac{\pi }{2}.$


Equation of line $OD$ is $\overrightarrow{r_1}=\hat{j}+\lambda (\frac{\hat{i}-\hat{j}}{2}+\frac{\hat{k}}{\sqrt{2}})$
and equation of line $AC$ is $\overrightarrow{r_2}=\overrightarrow{0}+\mu (\hat{i}+\hat{j})$

Now, shortest distance between $AC$ and $OD$ is
$\left|\frac{\{(\frac{\hat{i}-\hat{j}}{2}+\frac{\hat{k}}{\sqrt{2}})\times (\hat{i}+\hat{j}\left)\right\}\cdot (\overrightarrow{0}-\overrightarrow{j})}{|(\frac{\hat{i}-\hat{j}}{2}+\frac{\hat{k}}{\sqrt{2}})\times (\hat{i}+\hat{j}\left)\right|}\right|$
$=\frac{1}{2}$ unit