arthham

JEE Advanced 2026 · Three Dimensional Geometry

Let P be the plane such that it contains the straight line x−12=y−33=z+21\frac{x-1}{2} = \frac{y-3}{3} = \frac{z+2}{1} and is perpendicular to the plane x+2y+3z=4x + 2y + 3z = 4. Let P1P_1 be the plane which passes through the point (4,2,2)(4, 2, 2) and is parallel to P. Then which of the following statements is (are) TRUE?

Answers: A. The equation of the plane P is 7x−5y+z=−107x - 5y + z = -10 · D. The acute angle between the plane P and the plane 2x+2y+z=32x + 2y + z = 3 is cos⁡−1(133)\cos^{-1}\left(\frac{1}{3\sqrt{3}}\right)

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Given: P⊃L: x−12=y−33=z+21\text{Given: } P \supset L:\ \dfrac{x-1}{2} = \dfrac{y-3}{3} = \dfrac{z+2}{1} P⊥ plane x+2y+3z=4P \perp \text{ plane } x + 2y + 3z = 4 P1: through (4, 2, 2),  P1∥PP_1:\ \text{through } (4,\,2,\,2),\ \ P_1 \parallel P To find: which of (1), (2), (3), (4) are TRUE\text{To find: which of } (1),\ (2),\ (3),\ (4) \text{ are TRUE} Plan: 1. m=d⃗×n⃗⇒P\text{Plan: } 1.\ m = \vec{d} \times \vec{n} \Rightarrow P 2. P1 parallel through (4,2,2)2.\ P_1 \text{ parallel through } (4,2,2) 3. distances and angle between planes3.\ \text{distances and angle between planes} L: r⃗=(1, 3, −2)+λ(2, 3, 1)L:\ \vec{r} = (1,\,3,\,-2) + \lambda(2,\,3,\,1) Π: x+2y+3z=4,n⃗=(1, 2, 3)\Pi:\ x + 2y + 3z = 4,\qquad \vec{n} = (1,\,2,\,3) L⊂P⇒m⃗⋅d⃗=0L \subset P \Rightarrow \vec{m} \cdot \vec{d} = 0 P⊥Π⇒m⃗⋅n⃗=0P \perp \Pi \Rightarrow \vec{m} \cdot \vec{n} = 0 m⃗=d⃗×n⃗\vec{m} = \vec{d} \times \vec{n} m⃗=d⃗×n⃗=∣i^j^k^231123∣\vec{m} = \vec{d} \times \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 3 \end{vmatrix} =i^(9−2)−j^(6−1)+k^(4−3)= \hat{i}(9-2) - \hat{j}(6-1) + \hat{k}(4-3) m⃗=7i^−5j^+k^=(7, −5, 1)\vec{m} = 7\hat{i} - 5\hat{j} + \hat{k} = (7,\,-5,\,1) ∣m⃗∣=72+(−5)2+12=49+25+1=75=53|\vec{m}| = \sqrt{7^2 + (-5)^2 + 1^2} = \sqrt{49 + 25 + 1} = \sqrt{75} = 5\sqrt{3} P: 7x−5y+z=k,k=7(1)−5(3)+(−2)P:\ 7x - 5y + z = k,\qquad k = 7(1) - 5(3) + (-2) k=7−15−2=−10⇒P: 7x−5y+z=−10k = 7 - 15 - 2 = -10 \Rightarrow P:\ 7x - 5y + z = -10 ⇒ (1) is TRUE\Rightarrow\ (1)\ \text{is TRUE} P1: 7x−5y+z=c,c=7(4)−5(2)+2P_1:\ 7x - 5y + z = c,\qquad c = 7(4) - 5(2) + 2 c=28−10+2=20⇒P1: 7x−5y+z=20c = 28 - 10 + 2 = 20 \Rightarrow P_1:\ 7x - 5y + z = 20 d=∣c1−c2∣a2+b2+c2=∣−10−20∣53d = \dfrac{|c_1 - c_2|}{\sqrt{a^2+b^2+c^2}} = \dfrac{|-10 - 20|}{5\sqrt{3}} =3053=63=23= \dfrac{30}{5\sqrt{3}} = \dfrac{6}{\sqrt{3}} = 2\sqrt{3} d(P, P1)=23≠30 ⇒ (2) is FALSEd(P,\,P_1) = 2\sqrt{3} \neq 30\ \Rightarrow\ (2)\ \text{is FALSE} d(O, P)=∣−10∣75=1053=23=233d(O,\,P) = \dfrac{|-10|}{\sqrt{75}} = \dfrac{10}{5\sqrt{3}} = \dfrac{2}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3} 233≠23 ⇒ (3) is FALSE\dfrac{2\sqrt{3}}{3} \neq 2\sqrt{3}\ \Rightarrow\ (3)\ \text{is FALSE} cos⁡θ=∣m⃗⋅n⃗2∣∣m⃗∣ ∣n⃗2∣\cos\theta = \dfrac{|\vec{m} \cdot \vec{n}_2|}{|\vec{m}|\,|\vec{n}_2|} =∣7(2)−5(2)+1(1)∣(53)(3)=5153=133= \dfrac{|7(2) - 5(2) + 1(1)|}{(5\sqrt{3})(3)} = \dfrac{5}{15\sqrt{3}} = \dfrac{1}{3\sqrt{3}} θ=cos⁡−1 ⁣(133) ⇒ (4) is TRUE\theta = \cos^{-1}\!\left(\dfrac{1}{3\sqrt{3}}\right)\ \Rightarrow\ (4)\ \text{is TRUE} (1) ✓(2) ×(3) ×(4) ✓(1)\ \checkmark \qquad (2)\ \times \qquad (3)\ \times \qquad (4)\ \checkmark (1) and (4) are TRUE\boxed{(1)\ \text{and}\ (4)\ \text{are TRUE}}

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