JEE Advance - Mathematics (2018 - Paper 2 Offline - No. 12)
Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the line segment PQ is perpendicular to the plane x + y = 3 and the mid-point of PQ lies in the plane x + y = 3) lies on the Z-axis. Let the distance of P from the X-axis be 5. If R is the image of P in the XY-plane, then the length of PR is ...............
Answer
8
Explanation
Let P($$\alpha $$, $$\beta $$, $$\gamma $$) and R is image of P in the XY-plane.
$$ \therefore $$ R($$\alpha $$, $$\beta $$, $$-$$$$\gamma $$)
Also, Q is the image of P in the plane x + y = 3
$$ \therefore $$ $${{x - \alpha } \over 1} = {{y - \beta } \over 1} = {{z - \gamma } \over 0}$$
$$ = {{ - 2(\alpha + \beta - 3)} \over 2}$$
$$x = 3 - \beta ,\,y = 3 - \alpha ,\,z = \gamma $$
Since, Q is lies on Z-axis
$$ \therefore $$ $$\beta = 3,\,\alpha = 3,\,z = \gamma $$
$$ \therefore $$ $$P(3,3,\gamma )$$
Given, distance of P from X-axis be 5
$$ \therefore $$ $$5 = \sqrt {{3^2} + {\gamma ^2}} $$
$$25 - 9 = {\gamma ^2}$$
$$ \Rightarrow \gamma = \pm 4$$
Then, $$PR = |2\gamma | = |2 \times 4| = 8$$
$$ \therefore $$ R($$\alpha $$, $$\beta $$, $$-$$$$\gamma $$)
Also, Q is the image of P in the plane x + y = 3
$$ \therefore $$ $${{x - \alpha } \over 1} = {{y - \beta } \over 1} = {{z - \gamma } \over 0}$$
$$ = {{ - 2(\alpha + \beta - 3)} \over 2}$$
$$x = 3 - \beta ,\,y = 3 - \alpha ,\,z = \gamma $$
Since, Q is lies on Z-axis
$$ \therefore $$ $$\beta = 3,\,\alpha = 3,\,z = \gamma $$
$$ \therefore $$ $$P(3,3,\gamma )$$
Given, distance of P from X-axis be 5
$$ \therefore $$ $$5 = \sqrt {{3^2} + {\gamma ^2}} $$
$$25 - 9 = {\gamma ^2}$$
$$ \Rightarrow \gamma = \pm 4$$
Then, $$PR = |2\gamma | = |2 \times 4| = 8$$
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