JEE Advance - Mathematics (2022 - Paper 1 Online - No. 12)
Let $$S$$ be the reflection of a point $$Q$$ with respect to the plane given by
$$ \vec{r}=-(t+p) \hat{\imath}+t \hat{\jmath}+(1+p) \hat{k} $$
where $$t, p$$ are real parameters and $$\hat{\imath}, \hat{\jmath}, \hat{k}$$ are the unit vectors along the three positive coordinate axes. If the position vectors of $$Q$$ and $$S$$ are $$10 \hat{\imath}+15 \hat{\jmath}+20 \hat{k}$$ and $$\alpha \hat{\imath}+\beta \hat{\jmath}+\gamma \hat{k}$$ respectively, then which of the following is/are TRUE ?
$$ \vec{r}=-(t+p) \hat{\imath}+t \hat{\jmath}+(1+p) \hat{k} $$
where $$t, p$$ are real parameters and $$\hat{\imath}, \hat{\jmath}, \hat{k}$$ are the unit vectors along the three positive coordinate axes. If the position vectors of $$Q$$ and $$S$$ are $$10 \hat{\imath}+15 \hat{\jmath}+20 \hat{k}$$ and $$\alpha \hat{\imath}+\beta \hat{\jmath}+\gamma \hat{k}$$ respectively, then which of the following is/are TRUE ?
$$3(\alpha+\beta)=-101$$
$$3(\beta+\gamma)=-71$$
$$3(\gamma+\alpha)=-86$$
$$3(\alpha+\beta+\gamma)=-121$$
Explanation
Given : equation of plane
$$ \vec{r}=-(t+p) \hat{i}+t \hat{j}+(1+p) \hat{k} $$
on rearranging we get,
$$ \vec{r}=k+t+(-\hat{i}+\hat{j})+p(-\hat{i}+\hat{k}) $$
So, equation of plane in standard form is given by
$$ \begin{aligned} & {[\vec{r}-\hat{k}\,\,\,\,\,\,\,\,-\hat{i}+\hat{j}\,\,\,\,\,\,\,\,-\hat{i}+\hat{k}]=0} \\\\ & \Rightarrow\left[\begin{array}{rrr} x & y & z-1 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]=0 \\\\ & \Rightarrow x+y+z=1 .........(1) \end{aligned} $$
Now given co-ordinate of $Q=(10,15,20)$
And Co-ordinates of $S=(\alpha, \beta, \gamma)$
Now using the image formula of point and plane we get,
$$ \begin{aligned} & \frac{\alpha-10}{1}=\frac{\beta-15}{1}=\frac{\gamma-20}{1}=\frac{-2(10+15+20-1)}{3} \\\\ & \Rightarrow \alpha-10=\beta-15=\gamma-20=\frac{-88}{3} \\\\ & \Rightarrow \alpha=\frac{-58}{3}, \beta=\frac{-43}{3}, \gamma=\frac{-28}{3} \end{aligned} $$
Now solving all options.
$$ \begin{aligned} 3(\alpha+\beta) & =-101 \\\\ 3(\beta+\gamma) & =-71 \\\\ 3(\gamma+\alpha) & =-86 \\\\ 3(\alpha+\beta+\gamma) & =-129 \end{aligned} $$
$$ \vec{r}=-(t+p) \hat{i}+t \hat{j}+(1+p) \hat{k} $$
on rearranging we get,
$$ \vec{r}=k+t+(-\hat{i}+\hat{j})+p(-\hat{i}+\hat{k}) $$
So, equation of plane in standard form is given by
$$ \begin{aligned} & {[\vec{r}-\hat{k}\,\,\,\,\,\,\,\,-\hat{i}+\hat{j}\,\,\,\,\,\,\,\,-\hat{i}+\hat{k}]=0} \\\\ & \Rightarrow\left[\begin{array}{rrr} x & y & z-1 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]=0 \\\\ & \Rightarrow x+y+z=1 .........(1) \end{aligned} $$
Now given co-ordinate of $Q=(10,15,20)$
And Co-ordinates of $S=(\alpha, \beta, \gamma)$
Now using the image formula of point and plane we get,
$$ \begin{aligned} & \frac{\alpha-10}{1}=\frac{\beta-15}{1}=\frac{\gamma-20}{1}=\frac{-2(10+15+20-1)}{3} \\\\ & \Rightarrow \alpha-10=\beta-15=\gamma-20=\frac{-88}{3} \\\\ & \Rightarrow \alpha=\frac{-58}{3}, \beta=\frac{-43}{3}, \gamma=\frac{-28}{3} \end{aligned} $$
Now solving all options.
$$ \begin{aligned} 3(\alpha+\beta) & =-101 \\\\ 3(\beta+\gamma) & =-71 \\\\ 3(\gamma+\alpha) & =-86 \\\\ 3(\alpha+\beta+\gamma) & =-129 \end{aligned} $$
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