JEE Advance - Mathematics (2020 - Paper 2 Offline - No. 10)
Let $$\alpha $$2 + $$\beta $$2 + $$\gamma $$2 $$ \ne $$ 0 and $$\alpha $$ + $$\gamma $$ = 1. Suppose the point (3, 2, $$-$$1) is the mirror image of the point (1, 0, $$-$$1) with respect to the plane $$\alpha $$x + $$\beta $$y + $$\gamma $$z = $$\delta $$. Then which of the following statements is/are TRUE?
$$\alpha $$ + $$\beta $$ = 2
$$\delta $$ $$-$$ $$\gamma $$ = 3
$$\delta $$ + $$\beta $$ = 4
$$\alpha $$ + $$\beta $$ + $$\gamma $$ = $$\delta $$
Explanation
Since, the point A(3, 2, $$-$$1) is the mirror image of the point B(1, 0, $$-$$1) with respect to the plane $$\alpha $$x + $$\beta $$y + $$\gamma $$z = $$\delta $$, then
$${\alpha \over {3 - 1}} = {\beta \over {2 - 0}} = {\gamma \over {( - 1) - ( - 1)}}$$
$$ \Rightarrow {\alpha \over 1} = {\beta \over 1} = {\gamma \over 0} \Rightarrow \gamma = 0$$
it is given that $$\alpha + \gamma = 1 \Rightarrow \alpha = 1$$, so $$\beta = 1$$.
And, the mid-point of AB, M(2, 1, $$-$$1) lies on the given plane, so
$$2\alpha + \beta - \gamma = \delta $$
$$ \Rightarrow 2(1) + (1) - 0 = \delta \Rightarrow \delta = 3$$
$$ \therefore $$ $$\alpha + \beta = 2$$, $$\delta - \gamma = 0$$, $$\delta + \beta = 4$$.
$${\alpha \over {3 - 1}} = {\beta \over {2 - 0}} = {\gamma \over {( - 1) - ( - 1)}}$$
$$ \Rightarrow {\alpha \over 1} = {\beta \over 1} = {\gamma \over 0} \Rightarrow \gamma = 0$$
it is given that $$\alpha + \gamma = 1 \Rightarrow \alpha = 1$$, so $$\beta = 1$$.
And, the mid-point of AB, M(2, 1, $$-$$1) lies on the given plane, so
$$2\alpha + \beta - \gamma = \delta $$
$$ \Rightarrow 2(1) + (1) - 0 = \delta \Rightarrow \delta = 3$$
$$ \therefore $$ $$\alpha + \beta = 2$$, $$\delta - \gamma = 0$$, $$\delta + \beta = 4$$.
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