Find the equation of plane passing...

MathematicsEquation of a PlaneJEE Advanced 2004Moderate

Find the equation of plane passing through $(1, 1, 1)$ & parallel to the lines $L_{1}, L_{2}$ having direction ratios $(1, 0, - 1), (1, - 1, 0)$ . Find the volume of tetrahedron formed by origin and the points where these planes intersect the coordinate axes.

Answer:

x + y + z = 3

;

\frac{9}{2}

cubic units

Visualized Solution (Hindi)

Visualizing the Geometry

Given point: $P (1, 1, 1)$
Line $L_{1}$ direction ratios: $(1, 0, - 1)$
Line $L_{2}$ direction ratios: $(1, - 1, 0)$
Concept: The normal vector $n$ to the plane must be perpendicular to both $L_{1}$ and $L_{2}$ .

Finding the Normal Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Normal vector $n = L_{1} \times L_{2}$
$n = \hat{i} 11 \hat{j} 0 - 1 \hat{k} - 1 0$
Expanding the determinant:
$n = \hat{i} (0 - 1) - \hat{j} (0 - (- 1)) + \hat{k} (- 1 - 0)$
$n = - \hat{i} - \hat{j} - \hat{k}$

Equation of the Plane

Point-Normal form: $a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$
Substitute $a = - 1, b = - 1, c = - 1$ and $(x_{1}, y_{1}, z_{1}) = (1, 1, 1)$ :
$- 1 (x - 1) - 1 (y - 1) - 1 (z - 1) = 0$
Simplifying: $- (x - 1 + y - 1 + z - 1) = 0$
Final Plane Equation: $x + y + z = 3$

Finding the Intercepts

Convert to Intercept Form: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
Divide $x + y + z = 3$ by $3$ :
$\frac{x}{3} + \frac{y}{3} + \frac{z}{3} = 1$
Intercepts on axes: $A (3, 0, 0)$ , $B (0, 3, 0)$ , $C (0, 0, 3)$

Volume of the Tetrahedron

Vertices of tetrahedron: $O (0, 0, 0), A (3, 0, 0), B (0, 3, 0), C (0, 0, 3)$
Volume formula: $V = \frac{1}{6} ∣ a \cdot b \cdot c ∣$
Substitute $a = 3, b = 3, c = 3$ :
$V = \frac{1}{6} (3 \times 3 \times 3) = \frac{27}{6}$
Final Volume: $\frac{9}{2}$ cubic units

Summary and Key Takeaways

Key Takeaways:
Plane Equation: $x + y + z = 3$
Tetrahedron Volume: $\frac{9}{2}$ cubic units
Concept Check: Why is the normal perpendicular to parallel lines?
Next Challenge: Find the distance of the origin from this plane.

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Conceptually Similar Problems

MathematicsIntersection of a Line and a PlaneJEE Advanced 2012Moderate

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The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{3}$ from the point $(3, 1, - 1)$ is

(A)

5 x - 11 y + z = 17

(B)

2 x + y = 32 - 1

(C)

x + y + z = 3

(D)

x - 2 y = 1 - 2

Answer:A

MathematicsEquation of a PlaneJEE Advanced 2010Moderate

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Equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

(A)

x + 2 y - 2 z = 0

(B)

3 x + 2 y - 2 z = 0

(C)

x - 2 y + z = 0

(D)

5 x + 2 y - 4 z = 0

Answer:C

MathematicsEquation of a PlaneJEE Advanced 2005Moderate

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Find the equation of the plane containing the line $2 x - y + z - 3 = 0, 3 x + y + z = 5$ and at a distance of $\frac{1}{6}$ from the point $(2, 1, - 1)$ .

Answer:

62 x + 29 y + 19 z - 105 = 0

MathematicsEquation of a PlaneJEE Advanced 2003Moderate

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(i) Find the equation of the plane passing through the points $(2, 1, 0), (5, 0, 1)$ and $(4, 1, 1)$ . (ii) If $P$ is the point $(2, 1, 6)$ then find the point $Q$ such that $P Q$ is perpendicular to the plane in (i) and the midpoint of $P Q$ lies on it.

Answer:(i)

x + y - 2 z = 3

(ii)

Q (6, 5, - 2)

MathematicsEquation of a PlaneJEE Advanced 2006Moderate

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A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

(A)

0

(B)

1

(C)

2

(D)

22

Answer:D

MathematicsEquation of a PlaneJEE Main 2002Moderate

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A plane which passes through the point $(3, 2, 0)$ and the line $\frac{x - 4}{1} = \frac{y - 7}{5} = \frac{z - 4}{4}$ is

(A)

x - y + z = 1

(B)

x + y + z = 5

(C)

x + 2 y - z = 1

(D)

2 x - y + z = 5

Answer:A

MathematicsEquation of a PlaneJEE Main 2015Moderate

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The equation of the plane containing the line $2 x - 5 y + z = 3$ ; $x + y + 4 z = 5$ , and parallel to the plane, $x + 3 y + 6 z = 1$ , is:

(A)

x + 3 y + 6 z = 7

(B)

2 x + 6 y + 12 z = - 13

(C)

2 x + 6 y + 12 z = 13

(D)

x + 3 y + 6 z = - 7

Answer:C

MathematicsEquation of a PlaneJEE Advanced 2015Moderate

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In $R^{3}$ , consider the planes $P_{1} : y = 0$ and $P_{2} : x + z = 1$ . Let $P_{3}$ be the plane, different from $P_{1}$ and $P_{2}$ , which passes through the intersection of $P_{1}$ and $P_{2}$ . If the distance of the point $(0, 1, 0)$ from $P_{3}$ is 1 and the distance of a point $(α, β, γ)$ from $P_{3}$ is 2, then which of the following relations is (are) true?

* Multiple Correct Options

(A)

2 α + β + 2 γ + 2 = 0

(B)

2 α - β + 2 γ + 4 = 0

(C)

2 α + β - 2 γ - 10 = 0

(D)

2 α - β + 2 γ - 8 = 0

Answer:B, D

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Let $P$ be the image of the point $(3, 1, 7)$ with respect to the plane $x - y + z = 3$ . Then the equation of the plane passing through $P$ and containing the straight line $\frac{x}{1} = \frac{y}{2} = \frac{z}{1}$ is

(A)

x + y - 3 z = 0

(B)

3 x + z = 0

(C)

x - 4 y + 7 z = 0

(D)

2 x - y = 0

Answer:C

MathematicsEquation of a PlaneJEE Advanced 2013Moderate

View in:English Hindi

Consider the lines $L_{1} : \frac{x - 1}{2} = \frac{y}{- 1} = \frac{z + 3}{1}, L_{2} : \frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2}$ and the planes $P_{1} : 7 x + y + 2 z = 3, P_{2} : 3 x + 5 y - 6 z = 4$ . Let $a x + b y + cz = d$ be the equation of the plane passing through the point of intersection of lines $L_{1}$ and $L_{2}$ , and perpendicular to planes $P_{1}$ and $P_{2}$ . Match List I with List II:

List-I

(P)

(Q)

(R)

(S)

List-II

(1)

(2)

-3

(3)

(4)

-2

Answer:P → 3, Q → 2, R → 4, S → 1

Visualized Solution (Hindi)

Visualizing the Geometry

Finding the Normal Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Equation of the Plane

Finding the Intercepts

Volume of the Tetrahedron

Summary and Key Takeaways

Conceptually Similar Problems

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