JEE Main 2019 (9 January)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: The plane through the intersection of the planes $x + y + z = 1$ and $2 x + 3 y - z + 4 = 0$ and parallel to y-axis also passes through the point :

Q: Wait, why does the coefficient of $y$ become zero when the plane is parallel to the y-axis?

Think about the normal vector $\vec{n} = (A, B, C)$. The y-axis has a direction vector $\hat{j} = (0, 1, 0)$. If the plane is parallel to the y-axis, the normal vector must be perpendicular to the y-axis. Thus, $\vec{n} \cdot \hat{j} = 0$, which means $(A, B, C) \cdot (0, 1, 0) = B = 0$. Therefore, the coefficient of $y$ must be zero.

Select Answer:

Visualized Solution

Visualizing the Intersecting Planes

Given Planes:
$P_{1} : x + y + z - 1 = 0$
$P_{2} : 2 x + 3 y - z + 4 = 0$

The Family of Planes

Equation of the family of planes:
$P_{1} + λ P_{2} = 0$

Substituting the Equations

Substituting $P_{1}$ and $P_{2}$ :
$(x + y + z - 1) + λ (2 x + 3 y - z + 4) = 0$

Grouping the Variables

Rearranging terms by $x, y, z$ :
$(1 + 2 λ) x + (1 + 3 λ) y + (1 - λ) z + (4 λ - 1) = 0$

The Parallelism Condition

Condition: The plane is parallel to the y-axis.
Geometrically, the normal vector is perpendicular to the y-axis.

Applying the Condition

Dot product of normal and y-axis is zero.
Coefficient of $y$ must be zero:
$1 + 3 λ = 0$

Solving for $λ$

Solving the linear equation:
$3 λ = - 1$
$λ = - \frac{1}{3}$

Substituting $λ$ Back

Substitute $λ = - \frac{1}{3}$ into the grouped equation:
$(1 - \frac{2}{3}) x + 0 y + (1 + \frac{1}{3}) z + (- 1 - \frac{4}{3}) = 0$

Simplifying the Final Equation

Simplifying the fractions:
$\frac{1}{3} x + \frac{4}{3} z - \frac{7}{3} = 0$
Final Equation: $x + 4 z - 7 = 0$

Testing the Given Options

Check point $(3, 2, 1)$ :
Substitute $x = 3, z = 1$
$3 + 4 (1) - 7 = 3 + 4 - 7 = 0$
LHS = RHS

Final Conclusion

The point $(3, 2, 1)$ satisfies the equation.
Therefore, it lies on the required plane.
Final Answer: $(3, 2, 1)$

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

We are given two planes in 3D space:

P_{1} : x + y + z - 1 = 0

P_{2} : 2 x + 3 y - z + 4 = 0

Our objective is to determine the equation of a third plane that passes through the line of intersection of

P_{1}

and

P_{2}

while remaining parallel to the

y

-axis.

The Power of the Family of Planes

Any plane passing through the intersection of two given planes can be represented by the equation of the Family of Planes:

P_{1} + λ P_{2} = 0

Substituting the given expressions, we obtain:

(x + y + z - 1) + λ (2 x + 3 y - z + 4) = 0

By grouping the coefficients of

x

y

, and

z

, we rewrite the equation as:

(1 + 2 λ) x + (1 + 3 λ) y + (1 - λ) z + (4 λ - 1) = 0

The Constraint

Parallelism

A plane is parallel to the

y

-axis if and only if its normal vector

n

is perpendicular to the

y

-axis unit vector

\hat{j} = (0, 1, 0)

. This condition implies that the dot product

n \cdot \hat{j} = 0

The normal vector of our plane is

n = (1 + 2 λ, 1 + 3 λ, 1 - λ)

. Calculating the dot product with

\hat{j}

yields the coefficient of

y

1 + 3 λ = 0

The Final Calculation

Solving for the parameter

λ

, we find:

λ = - \frac{1}{3}

Substituting this value back into our general family equation:

(1 + 2 (- \frac{1}{3})) x + (1 + 3 (- \frac{1}{3})) y + (1 - (- \frac{1}{3})) z + (4 (- \frac{1}{3}) - 1) = 0

This simplifies to:

\frac{1}{3} x + 0 y + \frac{4}{3} z - \frac{7}{3} = 0

Multiplying the entire equation by 3, we arrive at the final equation of the plane:

x + 4 z - 7 = 0

Common Traps & FAQs

Wait, why does the coefficient of

y

become zero when the plane is parallel to the y-axis?

Think about the normal vector

n = (A, B, C)

. The y-axis has a direction vector

\hat{j} = (0, 1, 0)

. If the plane is parallel to the y-axis, the normal vector must be perpendicular to the y-axis. Thus,

n \cdot \hat{j} = 0

, which means

(A, B, C) \cdot (0, 1, 0) = B = 0

. Therefore, the coefficient of

y

must be zero.

Can we use the cross product of the normals of the two planes to find the line of intersection?

Yes, absolutely! The direction of the line of intersection is

v = n_{1} \times n_{2}

. However, using the 'Family of Planes' method is often faster for finding the equation of a specific plane passing through that line, especially when a condition like 'parallel to an axis' is given.

What if the plane was parallel to the x-axis instead?

The logic remains identical! If the plane is parallel to the x-axis, the normal vector must be perpendicular to the x-axis (direction

\hat{i} = (1, 0, 0)

). This would mean the coefficient of

x

(the first component of the normal vector) must be zero. You would set the coefficient of

x

to zero and solve for

λ

Animated Solution for Mathematics - Three Dimensional Geometry: The plane through the intersection of the planes $x + y + z = 1$ and $2 x + 3 y - z + 4 = 0$ and parallel to y-axis also passes through the point :

Select Answer:

Visualized Solution

Visualizing the Intersecting Planes

The Family of Planes

Substituting the Equations

Grouping the Variables

The Parallelism Condition

Applying the Condition

Solving for $λ$

Substituting $λ$ Back

Simplifying the Final Equation

Testing the Given Options

Final Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Power of the Family of Planes

The Constraint

The Final Calculation

Similar Questions

The equation of a plane containing the line of intersection of the planes $2 x - y - 4 = 0$ and $y + 2 z - 4 = 0$ and passing through the point $(1, 1, 0)$ is :

The plane passing through the point (4, -1, 2) and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point :

The plane passing through the points $(1, 2, 1)$ , $(2, 1, 2)$ and parallel to the line, $2 x = 3 y, z = 1$ also passes through the point:

The equation of the plane passing through the line of intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0$ and parallel to the $x$ -axis is:

Equation of a plane at a distance $\frac{2}{21}$ from the origin, which contains the line of intersection of the planes $x - y - z - 1 = 0$ and $2 x + y - 3 z + 4 = 0$ is :

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:

Let the equation of the plane, that passes through the point $(1, 4, - 3)$ and contains the line of intersection of the planes $3 x - 2 y + 4 z - 7 = 0$ and $x + 5 y - 2 z + 9 = 0$ , be $α x + β y + γ z + 3 = 0$ , then $α + β + γ$ is equal to :

A plane passing through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $π /4$ with the plane $y - z + 5 = 0$ , also passes through the point

The plane which bisects the line joining, the points $(4, - 2, 3)$ and $(2, 4, - 1)$ at right angles also passes through the point:

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

Select Answer:

Visualized Solution

Visualizing the Intersecting Planes

The Family of Planes

Substituting the Equations

Grouping the Variables

The Parallelism Condition

Applying the Condition

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ Back

Simplifying the Final Equation

Testing the Given Options

Final Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Power of the Family of Planes

The Constraint

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a plane containing the line of intersection of the planes 2x−y−4=0 and y+2z−4=0 and passing through the point (1,1,0) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The plane passing through the point (4, -1, 2) and parallel to the lines 3x+2​=−1y−2​=2z+1​ and 1x−2​=2y−3​=3z−4​ also passes through the point :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The plane passing through the points (1,2,1), (2,1,2) and parallel to the line, 2x=3y,z=1 also passes through the point:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the plane passing through the line of intersection of the planes r⋅(i^+j^​+k^)=1 and r⋅(2i^+3j^​−k^)+4=0 and parallel to the x-axis is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of a plane at a distance 212​​ from the origin, which contains the line of intersection of the planes x−y−z−1=0 and 2x+y−3z+4=0 is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the plane containing the line 2x−5y+z=3; x+y+4z=5, and parallel to the plane, x+3y+6z=1, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane passing through the points (0,−1,0) and (0,0,1) and making an angle π/4 with the plane y−z+5=0, also passes through the point

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The plane which bisects the line joining, the points (4,−2,3) and (2,4,−1) at right angles also passes through the point:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane which passes through the point (3,2,0) and the line 1x−4​=5y−7​=4z−4​ is

Solving for $λ$

Substituting $λ$ Back

The equation of a plane containing the line of intersection of the planes $2 x - y - 4 = 0$ and $y + 2 z - 4 = 0$ and passing through the point $(1, 1, 0)$ is :

The plane passing through the point (4, -1, 2) and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point :

The plane passing through the points $(1, 2, 1)$ , $(2, 1, 2)$ and parallel to the line, $2 x = 3 y, z = 1$ also passes through the point:

The equation of the plane passing through the line of intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0$ and parallel to the $x$ -axis is:

Equation of a plane at a distance $\frac{2}{21}$ from the origin, which contains the line of intersection of the planes $x - y - z - 1 = 0$ and $2 x + y - 3 z + 4 = 0$ is :

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 plane passing through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $π /4$ with the plane $y - z + 5 = 0$ , also passes through the point

The plane which bisects the line joining, the points $(4, - 2, 3)$ and $(2, 4, - 1)$ at right angles also passes through the point:

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