JEE Advance - Mathematics (2018 - Paper 1 Offline - No. 3)
Let P1 : 2x + y $$-$$ z = 3 and P2 : x + 2y + z = 2 be two planes. Then, which of the following statement(s) is(are) TRUE?
The line of intersection of P1 and P2 has direction ratios 1, 2, $$-$$1
The line $${{3x - 4} \over 9} = {{1 - 3y} \over 9} = {z \over 3}$$ is perpendicular to the line of intersection of P1 and P2
The acute angle between P1 and P2 is 60$$^\circ $$
If P3 is the plane passing through the point (4, 2, $$-$$2) and perpendicular to the line of intersection of P1 and P2, then the distance of the point (2, 1, 1) from the plane P3 is $${2 \over {\sqrt 3 }}$$
Explanation
We have,
and $$\eqalign{ & {P_1}:2x + y - z = 3 \cr & {P_2}:x + 2y + z = 2 \cr} $$
Here, $$\overrightarrow {{n_1}} = 2\widehat i + \widehat j - \widehat k$$
and $$\overrightarrow {{n_2}} = \widehat i + 2\widehat j + \widehat k$$
(a) Direction ratio of the line of intersection of P1 and P2 is $$\theta $$ $$\overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $$
i.e. $$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & { - 1} \cr 1 & 2 & 1 \cr } } \right|$$
$$ = (1 + 2)\widehat i - (2 + 1)\widehat j + (4 - 1)\widehat k$$
$$ = 3(\widehat i - \widehat j + \widehat k)$$
Hence, statement a is false.
(b) We have,
$${{3x - 4} \over 9} = {{1 - 3y} \over 9} = {z \over 3}$$
$$ \Rightarrow {{x - {4 \over 3}} \over 3} = {{\left( {y - {1 \over 3}} \right)} \over { - 3}} = {z \over 3}$$
This line is parallel to the line of intersection of P1 and P2.
Hence, statement (b) is false.
(c) Let acute angle between P1 and P2 be $$\theta $$.
We know that,
$$\cos \theta = {{\overrightarrow {{n_1}} .\overrightarrow {{n_2}} } \over {|\overrightarrow {{n_1}} ||\overrightarrow {{n_2}} |}}$$
$$ = {{(2\widehat i + \widehat j - \widehat k).(\widehat i + 2\widehat j + \widehat k)} \over {|2\widehat i + \widehat j - \widehat k||\widehat i + 2\widehat j + \widehat k|}}$$
$$ = {{2 + 2 - 1} \over {\sqrt 6 \times \sqrt 6 }} = {1 \over 2}$$
$$\theta = 60^\circ $$
Hence, statement (c) is true.
(d) Equation of plane passing through the point (4, 2, $$-$$2) and perpendicular to the line of intersection of P1 and P2 is
$$\eqalign{ & 3(x - 4) - 3(y - 2) + 3(z + 2) = 0 \cr & \Rightarrow 3x - 3y + 3z - 12 + 6 + 6 = 0 \cr} $$
$$ \Rightarrow 3x - 3y + 3z - 12 + 6 + 6 = 0$$
$$ \Rightarrow x - y + z = 0$$
Now, distance of the point (2, 1, 1) from the plane x $$-$$ y + z = 0 is
$$D = \left| {{{2 - 1 + 1} \over {\sqrt {1 + 1 + 1} }}} \right| = {2 \over {\sqrt 3 }}$$
Hence, statement (d) is true.
and $$\eqalign{ & {P_1}:2x + y - z = 3 \cr & {P_2}:x + 2y + z = 2 \cr} $$
Here, $$\overrightarrow {{n_1}} = 2\widehat i + \widehat j - \widehat k$$
and $$\overrightarrow {{n_2}} = \widehat i + 2\widehat j + \widehat k$$
(a) Direction ratio of the line of intersection of P1 and P2 is $$\theta $$ $$\overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $$
i.e. $$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & { - 1} \cr 1 & 2 & 1 \cr } } \right|$$
$$ = (1 + 2)\widehat i - (2 + 1)\widehat j + (4 - 1)\widehat k$$
$$ = 3(\widehat i - \widehat j + \widehat k)$$
Hence, statement a is false.
(b) We have,
$${{3x - 4} \over 9} = {{1 - 3y} \over 9} = {z \over 3}$$
$$ \Rightarrow {{x - {4 \over 3}} \over 3} = {{\left( {y - {1 \over 3}} \right)} \over { - 3}} = {z \over 3}$$
This line is parallel to the line of intersection of P1 and P2.
Hence, statement (b) is false.
(c) Let acute angle between P1 and P2 be $$\theta $$.
We know that,
$$\cos \theta = {{\overrightarrow {{n_1}} .\overrightarrow {{n_2}} } \over {|\overrightarrow {{n_1}} ||\overrightarrow {{n_2}} |}}$$
$$ = {{(2\widehat i + \widehat j - \widehat k).(\widehat i + 2\widehat j + \widehat k)} \over {|2\widehat i + \widehat j - \widehat k||\widehat i + 2\widehat j + \widehat k|}}$$
$$ = {{2 + 2 - 1} \over {\sqrt 6 \times \sqrt 6 }} = {1 \over 2}$$
$$\theta = 60^\circ $$
Hence, statement (c) is true.
(d) Equation of plane passing through the point (4, 2, $$-$$2) and perpendicular to the line of intersection of P1 and P2 is
$$\eqalign{ & 3(x - 4) - 3(y - 2) + 3(z + 2) = 0 \cr & \Rightarrow 3x - 3y + 3z - 12 + 6 + 6 = 0 \cr} $$
$$ \Rightarrow 3x - 3y + 3z - 12 + 6 + 6 = 0$$
$$ \Rightarrow x - y + z = 0$$
Now, distance of the point (2, 1, 1) from the plane x $$-$$ y + z = 0 is
$$D = \left| {{{2 - 1 + 1} \over {\sqrt {1 + 1 + 1} }}} \right| = {2 \over {\sqrt 3 }}$$
Hence, statement (d) is true.
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