The d.r. of normal to the plane through $(1, 0, 0), (0, 1, 0)$ which makes an angle $\pi/4$ with plane $x + y = 3$ are Options: (A) $1, \sqrt{2}, 1$ (B) $1, 1, \sqrt{2}$ (C) $1, 1, 2$ (D) $\sqrt{2}, 1, 1$

### Analyzing the Setup To define a plane in three-dimensional space, we require a point on the plane and a normal vector. We are given two points, $A(1, 0, 0)$ and $B(0, 1, 0)$, which lie on the plane. The general equation of a plane passing through $(x_1, y_1, z_1)$ with normal vector $(a, b, c)$ is: $$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$$ Substituting point $A(1, 0, 0)$ into this equation, we obtain the skeleton of our plane: $$a(x - 1) + by + cz = 0$$ ### Applying Constraints We must satisfy the condition that point $B(0, 1, 0)$ also lies on the plane. Substituting these coordinates into our equation yields: $$a(0 - 1) + b(1) + c(0) = 0$$ $$-a + b = 0 \Rightarrow b = a$$ Consequently, our normal vector $\vec{n_1} = (a, b, c)$ simplifies to $\vec{n_1} = (a, a, c)$. ### The Angle Condition We are given that our plane makes an angle of $\pi/4$ with the plane $x + y = 3$. The normal vector of this second plane is $\vec{n_2} = (1, 1, 0)$. The angle $\theta$ between two planes is defined by the angle between their normal vectors using the dot product formula: $$\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$$ Substituting $\theta = \pi/4$, $\vec{n_1} = (a, a, c)$, and $\vec{n_2} = (1, 1, 0)$, we get: $$\cos\left(\frac{\pi}{4}\right) = \frac{|a(1) + a(1) + c(0)|}{\sqrt{a^2 + a^2 + c^2} \sqrt{1^2 + 1^2 + 0^2}}$$ ### Final Calculation Simplifying the expression above, we have: $$\frac{1}{\sqrt{2}} = \frac{|2a|}{\sqrt{2a^2 + c^2} \sqrt{2}}$$ Canceling $\sqrt{2}$ from both sides and squaring the remaining terms results in: $$2a^2 + c^2 = (2a)^2$$ $$2a^2 + c^2 = 4a^2 \Rightarrow c^2 = 2a^2 \Rightarrow c = \sqrt{2}a$$ Thus, the normal vector is $(a, a, \sqrt{2}a)$. By setting $a=1$, we find the direction ratios of the normal to be **$(1, 1, \sqrt{2})$**.

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Animated Solution for Mathematics - Three Dimensional Geometry: The d.r. of normal to the plane through $(1, 0, 0), (0, 1, 0)$ which makes an angle $π /4$ with plane $x + y = 3$ are

Q: Why can we just divide by $a$ at the end? Doesn't that change the plane?

Not at all! Remember, the direction ratios $(a, b, c)$ define the *direction* of the normal, not its length. A normal vector $(1, 1, \sqrt{2})$ is parallel to $(2, 2, 2\sqrt{2})$. They represent the exact same orientation in space. Dividing by $a$ just gives us the simplest representation of that direction. You are essentially normalizing the vector!

Q: What if the angle was $0$?

If the angle were $0$, the planes would be parallel. Their normal vectors would be parallel, meaning their direction ratios would be proportional. You would find that the dot product formula gives $\cos(0) = 1$, leading to a scenario where the normal vectors point in the same direction. It is the same logic, just a different geometric outcome!

Select Answer:

Progressive Hints

Hint 1

Visualize the plane as a sheet of paper. If it passes through two points, the line segment connecting them lies entirely on that sheet. Your normal vector must be perpendicular to this line segment. Can you express this constraint mathematically?

Hint 2

The angle between two planes is defined by the angle between their normal vectors. Recall the dot product formula:

cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣}

. You have one normal vector from the given plane

x + y = 3

. Can you define the other in terms of variables?

Hint 3

Start with the general equation of a plane passing through

(1, 0, 0)

a (x - 1) + b y + cz = 0

. Since it also passes through

(0, 1, 0)

, substitute these coordinates to find a relationship between

a

and

b

. Then, use the angle condition

cos (π /4) = 1/ 2

to solve for

c

in terms of

a

Visualized Solution

Visualizing Points $A$ and $B$

Given points: $A (1, 0, 0)$ and $B (0, 1, 0)$
Objective: Find direction ratios $(a, b, c)$ of the required plane's normal.

Equation of Plane through $A (1, 0, 0)$

General equation of a plane through $(x_{1}, y_{1}, z_{1})$ :
$a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$

Substituting Point $A$

Substitute $A (1, 0, 0)$ :
$a (x - 1) + b (y - 0) + c (z - 0) = 0$

Simplified Plane Equation

Simplify the equation:
$a (x - 1) + b y + cz = 0$

Substituting Point $B (0, 1, 0)$

Substitute $B (0, 1, 0)$ into the equation:
$a (0 - 1) + b (1) + c (0) = 0$

Finding Relation $b = a$

$- a + b = 0 ⟹ b = a$
Updated normal direction ratios: $(a, a, c)$

Given Plane $x + y = 3$

Given plane: $x + y = 3$
Normal vector direction ratios: $(1, 1, 0)$

Angle Between Planes $θ = \frac{π}{4}$

Angle between planes formula:
$cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣}$

Applying $cos θ$ Formula

Substitute known values:
$cos (\frac{π}{4}) = \frac{∣ a ( 1 ) + a ( 1 ) + c ( 0 ) ∣}{a ^{2} + a ^{2} + c ^{2} 1 ^{2} + 1 ^{2} + 0 ^{2}}$

Simplifying the Expression

Simplify the expression:
$\frac{1}{2} = \frac{∣2 a ∣}{2 a ^{2} + c ^{2} 2}$

Squaring Both Sides

Cancel $2$ and square both sides:
$1 = \frac{4 a ^{2}}{2 a ^{2} + c ^{2}}$
$2 a^{2} + c^{2} = 4 a^{2}$

Solving for $c$

Solve for $c$ :
$c^{2} = 2 a^{2} ⟹ c = 2 a$

Final Direction Ratios

Final direction ratios: $(a, a, 2 a)$
Divide by $a$ : $(1, 1, 2)$

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The Sigma Insight: Angle Between Two Planes

Solution Diagram

Analyzing the Setup

To define a plane in three-dimensional space, we require a point on the plane and a normal vector. We are given two points,

A (1, 0, 0)

and

B (0, 1, 0)

, which lie on the plane.

The general equation of a plane passing through

(x_{1}, y_{1}, z_{1})

with normal vector

(a, b, c)

is:

a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0

Substituting point

A (1, 0, 0)

into this equation, we obtain the skeleton of our plane:

a (x - 1) + b y + cz = 0

Applying Constraints

We must satisfy the condition that point

B (0, 1, 0)

also lies on the plane. Substituting these coordinates into our equation yields:

a (0 - 1) + b (1) + c (0) = 0

- a + b = 0 \Rightarrow b = a

Consequently, our normal vector

n_{1} = (a, b, c)

simplifies to

n_{1} = (a, a, c)

The Angle Condition

We are given that our plane makes an angle of

π /4

with the plane

x + y = 3

. The normal vector of this second plane is

n_{2} = (1, 1, 0)

The angle

θ

between two planes is defined by the angle between their normal vectors using the dot product formula:

cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣}

Substituting

θ = π /4

n_{1} = (a, a, c)

, and

n_{2} = (1, 1, 0)

, we get:

cos (\frac{π}{4}) = \frac{∣ a ( 1 ) + a ( 1 ) + c ( 0 ) ∣}{a ^{2} + a ^{2} + c ^{2} 1 ^{2} + 1 ^{2} + 0 ^{2}}

Final Calculation

Simplifying the expression above, we have:

\frac{1}{2} = \frac{∣2 a ∣}{2 a ^{2} + c ^{2} 2}

Canceling

2

from both sides and squaring the remaining terms results in:

2 a^{2} + c^{2} = (2 a)^{2}

2 a^{2} + c^{2} = 4 a^{2} \Rightarrow c^{2} = 2 a^{2} \Rightarrow c = 2 a

Thus, the normal vector is

(a, a, 2 a)

. By setting

a = 1

, we find the direction ratios of the normal to be $(1, 1, 2)$ .

Common Traps & FAQs

Wait, why do we use the normal vector to find the angle between planes? Isn't the angle between the planes themselves?

That is a brilliant question! Think of a plane as an infinite sheet. Measuring the angle between two infinite sheets is tricky. However, the orientation of a plane is uniquely defined by its normal vector—the stick poking straight out of it. If you tilt the plane, the normal tilts by the exact same amount. Thus, the angle between the planes is identical to the angle between their normals. It is a beautiful geometric shortcut!

Why can we just divide by

a

at the end? Doesn't that change the plane?

Not at all! Remember, the direction ratios

(a, b, c)

define the *direction* of the normal, not its length. A normal vector

(1, 1, 2)

is parallel to

(2, 2, 22)

. They represent the exact same orientation in space. Dividing by

a

just gives us the simplest representation of that direction. You are essentially normalizing the vector!

What if the angle was

0

If the angle were

0

, the planes would be parallel. Their normal vectors would be parallel, meaning their direction ratios would be proportional. You would find that the dot product formula gives

cos (0) = 1

, leading to a scenario where the normal vectors point in the same direction. It is the same logic, just a different geometric outcome!

Animated Solution for Mathematics - Three Dimensional Geometry: The d.r. of normal to the plane through $(1, 0, 0), (0, 1, 0)$ which makes an angle $π /4$ with plane $x + y = 3$ are

Select Answer:

Visualized Solution

Visualizing Points $A$ and $B$

Equation of Plane through $A (1, 0, 0)$

Substituting Point $A$

Simplified Plane Equation

Substituting Point $B (0, 1, 0)$

Finding Relation $b = a$

Given Plane $x + y = 3$

Angle Between Planes $θ = \frac{π}{4}$

Applying $cos θ$ Formula

Simplifying the Expression

Squaring Both Sides

Solving for $c$

Final Direction Ratios

The Sigma Insight: Angle Between Two Planes

Analyzing the Setup

Applying Constraints

The Angle Condition

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Three Dimensional Geometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The d.r. of normal to the plane through (1,0,0),(0,1,0) which makes an angle π/4 with plane x+y=3 are

Select Answer:

Visualized Solution

Visualizing Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Equation of Plane through .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A(1,0,0)

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

Simplified Plane Equation

Substituting Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B(0,1,0)

Finding Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b=a

Given Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x+y=3

Angle Between Planes .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ=4π​

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosθ Formula

Simplifying the Expression

Squaring Both Sides

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

Final Direction Ratios

The Sigma Insight: Angle Between Two Planes

Analyzing the Setup

Applying Constraints

The Angle Condition

Final Calculation

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Animated Solution for Mathematics - Three Dimensional Geometry: The d.r. of normal to the plane through $(1, 0, 0), (0, 1, 0)$ which makes an angle $π /4$ with plane $x + y = 3$ are

Visualizing Points $A$ and $B$

Equation of Plane through $A (1, 0, 0)$

Substituting Point $A$

Substituting Point $B (0, 1, 0)$

Finding Relation $b = a$

Given Plane $x + y = 3$

Angle Between Planes $θ = \frac{π}{4}$

Applying $cos θ$ Formula

Solving for $c$

Let the points on the plane $P$ be equidistant from the points $(- 4, 2, 1)$ and $(2, - 2, 3)$ . Then the acute angle between the plane $P$ and the plane $2 x + y + 3 z = 1$ is

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