The direction ratios of normal to the plane through the points (0, -1, 0) and (0, 0, 1) and making a

Q: Wait, why can't we just take the cross product of the two points to find the normal vector?

That is a very common trap! The two points $A$ and $B$ define a line, not a plane. A single line does not uniquely determine a normal vector because a plane can rotate around that line like a door on a hinge. We need the angle condition to 'lock' the plane into its specific orientation.

Q: Why do we get $\pm \sqrt{2}b$ for $a$? Does the sign matter?

Great observation! The $\pm$ sign appears because of the modulus in the angle formula. Geometrically, it means there are two possible planes that satisfy the condition—one tilted one way and one tilted the other way, both making a $45^\circ$ angle with the reference plane. Both are valid solutions!

Q: Does the value of $d$ matter in the final answer?

Not at all! Remember that the equation of a plane $ax + by + cz + d = 0$ defines the plane's orientation via the normal vector $(a, b, c)$. The constant $d$ only shifts the plane parallel to itself. Since we are looking for direction ratios, we only care about the ratios of $a, b,$ and $c$. We can set $d$ to any non-zero value (or just use it as a placeholder) without affecting the direction ratios.

Question

The direction ratios of normal to the plane through the points $(0, -1, 0)$ and $(0, 0, 1)$ and making an angle $\pi/4$ with the plane $y - z + 5 = 0$ are:

Options:

(A) $2\sqrt{3}, 1, -1$

(B) $2, \sqrt{2}, -\sqrt{2}$

(C) $2, -1, 1$

(D) $\sqrt{2}, 1, -1$

Sigma Percentile · Accepted Answer

### The Geometry of the Invisible Plane

Welcome, future engineer. Today, we are not just solving a problem; we are sculpting in three-dimensional space. Imagine you are standing in a vast, empty room with two points, $A(0, -1, 0)$ and $B(0, 0, 1)$, floating in the air.

You are tasked with creating a flat, infinite sheet—a plane—that passes through both of these points. This plane must be tilted at a specific angle, $	heta = \frac{\pi}{4}$, relative to a reference plane defined by $y - z + 5 = 0$. This is the essence of JEE Advanced geometry: taking abstract constraints and turning them into a concrete, solvable reality.

### Phase 1: The DNA of the Plane

Every plane has a unique 'DNA'—its normal vector. If you know the normal vector $\vec{n}_1 = (a, b, c)$, you know everything about the plane's orientation. The equation of our plane is $ax + by + cz + d = 0$.

We know the plane passes through $A(0, -1, 0)$. Substituting these coordinates into our general equation:
$$a(0) + b(-1) + c(0) + d = 0 \Rightarrow -b + d = 0 \Rightarrow b = d$$

Similarly, the plane passes through $B(0, 0, 1)$. Substituting these coordinates:
$$a(0) + b(0) + c(1) + d = 0 \Rightarrow c + d = 0 \Rightarrow c = -d$$

Since $b = d$, we substitute $b$ for $d$ to find $c = -b$. Our normal vector $\vec{n}_1 = (a, b, c)$ transforms into $\vec{n}_1 = (a, b, -b)$. We have successfully reduced three unknowns down to two.

### Phase 2: The Bridge of Angles

Now, we look at the reference plane: $y - z + 5 = 0$. The normal vector to this plane, $\vec{n}_2$, is composed of the coefficients of $x, y,$ and $z$. Since there is no $x$ term, $\vec{n}_2 = (0, 1, -1)$.

In 3D geometry, the angle between two planes is identical to the angle between their normal vectors. We use the dot product formula:
$$\cos 	heta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$$

Given $	heta = \frac{\pi}{4}$, we know $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. This is the bridge that connects our unknown plane to the known reference plane.

### Phase 3: The Algebraic Resolution

The dot product $\vec{n}_1 \cdot \vec{n}_2$ is calculated as:
$$\vec{n}_1 \cdot \vec{n}_2 = (a)(0) + (b)(1) + (-b)(-1) = 2b$$

The magnitudes are:
$$|\vec{n}_1| = \sqrt{a^2 + b^2 + (-b)^2} = \sqrt{a^2 + 2b^2}$$
$$|\vec{n}_2| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$$

Plugging these into our formula:
$$\frac{1}{\sqrt{2}} = \frac{|2b|}{\sqrt{a^2 + 2b^2} \cdot \sqrt{2}}$$

The $\sqrt{2}$ in the denominator on both sides cancels out perfectly. We are left with:
$$1 = \frac{|2b|}{\sqrt{a^2 + 2b^2}}$$

Squaring both sides to eliminate the square root and the modulus:
$$a^2 + 2b^2 = 4b^2 \Rightarrow a^2 = 2b^2 \Rightarrow a = \pm \sqrt{2}b$$

### Conclusion: The Final Reveal

We have found the relationship between $a$ and $b$. If we set $b = 1$, then $a = \pm \sqrt{2}$. Our normal vector $\vec{n}_1$ becomes $(\pm \sqrt{2}, 1, -1)$.

If we multiply this by $\sqrt{2}$ to match standard coordinate representations, we get **$(2, \sqrt{2}, -\sqrt{2})$**. This is the direction ratio of the normal to our plane. You have navigated the 3D space, applied the constraints, and solved the system.

Animated Solution for Mathematics - Three Dimensional Geometry: The direction ratios of normal to the plane through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $π /4$ with the plane $y - z + 5 = 0$ are:

Select Answer:

Visualized Solution

Visualizing the 3D Setup

Equation of the Required Plane

Substituting Point $A$

Simplifying for Point $A$

Substituting Point $B$

Simplifying for Point $B$

The Normal Vector $n_{1}$

Normal of the Reference Plane

Angle Between Two Planes

Substituting into the Formula

Simplifying the Expression

Solving for $a$

Finding $a$ in terms of $b$

Final Direction Ratios

The Sigma Insight: Equation of a Plane

The Geometry of the Invisible Plane

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

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

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

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.

The 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 vector $v$ in the first octant is inclined to the $x$ -axis at $6 0^{\circ}$ , to the $y$ -axis at $4 5^{\circ}$ and to the $z$ -axis at an acute angle. If a plane passing through the points $(2, - 1, 1)$ and $(a, b, c)$ , is normal to $v$ , then

The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is :

The distance of the point $(1, 3, - 7)$ from the plane passing through the point $(1, - 1, - 1)$ , having normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$ , 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 :

Let the plane passing through the point $(- 1, 0, - 2)$ and perpendicular to each of the planes $2 x + y - z = 2$ and $x - y - z = 3$ be $a x + b y + cz + 8 = 0$ . Then the value of $a + b + c$ is equal to:

Let a plane $P$ contain two lines $r = \hat{i} + λ (\hat{i} + \hat{j}), λ \in R$ and $r = - \hat{j} + μ (\hat{j} - \hat{k}), μ \in R$ . If $Q (α, β, γ)$ is the foot of the perpendicular drawn from the point $M (1, 0, 1)$ to $P$ , then $3 (α + β + γ)$ equals