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 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

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

3D Geometry PYQ — JEE Advanced 2018 · Shift 1 · Q21 with Solution

View written solutionFree

Correct answer: C, D

Normals of the planes

Given $P_1: 2x+y-z=3, \qquad P_2: x+2y+z=2$

Their normal vectors are $\vec n_1=(2,1,-1), \qquad \vec n_2=(1,2,1).$

Direction ratios of the line of intersection

The line of intersection of two planes is parallel to $\vec n_1\times \vec n_2.$

Compute:

\begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix}$$ $$=\hat i(1\cdot 1-(-1)\cdot 2)-\hat j(2\cdot 1-(-1)\cdot 1)+\hat k(2\cdot 2-1\cdot 1)$$ $$=\hat i(3)-\hat j(3)+\hat k(3)=(3,-3,3).$$ So direction ratios are proportional to $$(1,-1,1).$$ ### Checking option A Option A says direction ratios are $(1,2,-1)$, which is **not** proportional to $(1,-1,1)$. So, **A is false**. --- 3. **Check whether the given line is perpendicular to the intersection line** Given line: $$\frac{3x-4}{9}=\frac{1-3y}{9}=\frac z3$$ Let the common value be $t$. Then $$3x-4=9t \Rightarrow x=3t+\frac43,$$ $$1-3y=9t \Rightarrow y=\frac13-3t,$$ $$z=3t.$$ Hence its direction vector is $$(3,-3,3)\propto (1,-1,1).$$ But the intersection line also has direction vector proportional to $(1,-1,1)$. So the given line is **parallel**, not perpendicular. ### Checking option B Therefore, **B is false**. --- 4. **Angle between the planes** The acute angle between two planes equals the acute angle between their normals. So $$\cos\theta=\frac{|\vec n_1\cdot \vec n_2|}{|\vec n_1||\vec n_2|}.$$ Now, $$\vec n_1\cdot \vec n_2=2\cdot 1+1\cdot 2+(-1)\cdot 1=2+2-1=3.$$ Also, $$|\vec n_1|=\sqrt{2^2+1^2+(-1)^2}=\sqrt6,$$ $$|\vec n_2|=\sqrt{1^2+2^2+1^2}=\sqrt6.$$ Thus $$\cos\theta=\frac{3}{\sqrt6\cdot \sqrt6}=\frac{3}{6}=\frac12.$$ So $$\theta=60^\circ.$$ ### Checking option C **C is true**. --- 5. **Plane perpendicular to the line of intersection** If a plane is perpendicular to a line, then the line's direction vector is a normal to the plane. The line of intersection has direction vector proportional to $(1,-1,1)$. So plane $P_3$ has normal vector $(1,-1,1)$ and passes through $(4,2,-2)$. Hence its equation is $$1(x-4)-1(y-2)+1(z+2)=0$$ $$x-y+z=0.$$ --- 6. **Distance of point $(2,1,1)$ from $P_3$** Distance from point $(x_1,y_1,z_1)$ to plane $ax+by+cz+d=0$ is $$\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}.$$ For plane $$x-y+z=0,$$ we have $a=1, b=-1, c=1, d=0$. For point $(2,1,1)$, $$\text{distance}=\frac{|2-1+1|}{\sqrt{1^2+(-1)^2+1^2}}= rac{2}{\sqrt3}.$$ ### Checking option D **D is true**. --- 7. **Final conclusion** The true statements are: $$\boxed{C, D}$$ This matches the stored correct answer.

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View written solutionFree

Correct answer: C, D

Normals of the planes

Given $P_1: 2x+y-z=3, \qquad P_2: x+2y+z=2$

Their normal vectors are $\vec n_1=(2,1,-1), \qquad \vec n_2=(1,2,1).$

Direction ratios of the line of intersection

The line of intersection of two planes is parallel to $\vec n_1\times \vec n_2.$

Compute:

\begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix}$$ $$=\hat i(1\cdot 1-(-1)\cdot 2)-\hat j(2\cdot 1-(-1)\cdot 1)+\hat k(2\cdot 2-1\cdot 1)$$ $$=\hat i(3)-\hat j(3)+\hat k(3)=(3,-3,3).$$ So direction ratios are proportional to $$(1,-1,1).$$ ### Checking option A Option A says direction ratios are $(1,2,-1)$, which is **not** proportional to $(1,-1,1)$. So, **A is false**. --- 3. **Check whether the given line is perpendicular to the intersection line** Given line: $$\frac{3x-4}{9}=\frac{1-3y}{9}=\frac z3$$ Let the common value be $t$. Then $$3x-4=9t \Rightarrow x=3t+\frac43,$$ $$1-3y=9t \Rightarrow y=\frac13-3t,$$ $$z=3t.$$ Hence its direction vector is $$(3,-3,3)\propto (1,-1,1).$$ But the intersection line also has direction vector proportional to $(1,-1,1)$. So the given line is **parallel**, not perpendicular. ### Checking option B Therefore, **B is false**. --- 4. **Angle between the planes** The acute angle between two planes equals the acute angle between their normals. So $$\cos\theta=\frac{|\vec n_1\cdot \vec n_2|}{|\vec n_1||\vec n_2|}.$$ Now, $$\vec n_1\cdot \vec n_2=2\cdot 1+1\cdot 2+(-1)\cdot 1=2+2-1=3.$$ Also, $$|\vec n_1|=\sqrt{2^2+1^2+(-1)^2}=\sqrt6,$$ $$|\vec n_2|=\sqrt{1^2+2^2+1^2}=\sqrt6.$$ Thus $$\cos\theta=\frac{3}{\sqrt6\cdot \sqrt6}=\frac{3}{6}=\frac12.$$ So $$\theta=60^\circ.$$ ### Checking option C **C is true**. --- 5. **Plane perpendicular to the line of intersection** If a plane is perpendicular to a line, then the line's direction vector is a normal to the plane. The line of intersection has direction vector proportional to $(1,-1,1)$. So plane $P_3$ has normal vector $(1,-1,1)$ and passes through $(4,2,-2)$. Hence its equation is $$1(x-4)-1(y-2)+1(z+2)=0$$ $$x-y+z=0.$$ --- 6. **Distance of point $(2,1,1)$ from $P_3$** Distance from point $(x_1,y_1,z_1)$ to plane $ax+by+cz+d=0$ is $$\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}.$$ For plane $$x-y+z=0,$$ we have $a=1, b=-1, c=1, d=0$. For point $(2,1,1)$, $$\text{distance}=\frac{|2-1+1|}{\sqrt{1^2+(-1)^2+1^2}}= rac{2}{\sqrt3}.$$ ### Checking option D **D is true**. --- 7. **Final conclusion** The true statements are: $$\boxed{C, D}$$ This matches the stored correct answer.

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