If λ1

Q: Why do we use $\cos \theta$ when the problem gives $\sin \theta$?

The angle between two planes is defined as the angle between their normal vectors. The standard dot product formula naturally yields $\cos \theta$. Converting $\sin \theta$ to $\cos \theta$ is a necessary step to align our given data with the geometric formula.

Q: Why is there an absolute value in the dot product formula?

The angle between two planes is conventionally defined as the acute angle between them. The dot product can be negative depending on the direction of the normal vectors, so the absolute value ensures we are calculating the cosine of the acute angle.

Q: Why do we reject one of the roots or how do we know which is $\lambda_1$?

We don't reject either root! We are given the condition $\lambda_1 < \lambda_2$. We simply solve the quadratic, find the two values, and assign the smaller one to $\lambda_1$ and the larger one to $\lambda_2$ to satisfy the problem's constraint.

Question

If $\lambda_1 < \lambda_2$ are two values of $\lambda$ such that the angle between the planes $P_1: \vec{r} \cdot (3\hat{i} - 5\hat{j} + \hat{k}) = 7$ and $P_2: \vec{r} \cdot (\lambda\hat{i} + \hat{j} - 3\hat{k}) = 9$ is $\sin^{-1}(\frac{2\sqrt{6}}{5})$, then the square of the length of perpendicular from the point $(38\lambda_1, 10\lambda_2, 2)$ to the plane $P_1$ is ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The geometry of planes allows us to navigate three-dimensional space using normal vectors. We are given two planes, $P_1$ and $P_2$, defined by their normal vectors.

For $P_1: \vec{r} \cdot (3\hat{i} - 5\hat{j} + \hat{k}) = 7$, the normal vector is $\vec{n}_1 = 3\hat{i} - 5\hat{j} + \hat{k}$.

For $P_2: \vec{r} \cdot (\lambda\hat{i} + \hat{j} - 3\hat{k}) = 9$, the normal vector is $\vec{n}_2 = \lambda\hat{i} + \hat{j} - 3\hat{k}$. Our primary objective is to determine the value of $\lambda$.

### The Angle Trap

We are given the angle $	heta$ between the planes as $\sin 	heta = \frac{2\sqrt{6}}{5}$. Because the angle between planes is defined by the cosine of the angle between their normal vectors, we must convert this value.

Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$, we calculate:

$$\cos^2 	heta = 1 - \left( \frac{2\sqrt{6}}{5} ight)^2 = 1 - \frac{24}{25} = \frac{1}{25}$$

Thus, $\cos 	heta = \frac{1}{5}$. This value serves as the bridge between our geometric constraints and the algebraic solution.

### The Algebraic Battle

We utilize the dot product formula for the angle between two planes:

$$\cos 	heta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$$

Substituting our known values, we obtain:

$$\frac{1}{5} = \frac{|(3)(\lambda) + (-5)(1) + (1)(-3)|}{\sqrt{3^2 + (-5)^2 + 1^2} \sqrt{\lambda^2 + 1^2 + (-3)^2}}$$

Simplifying the expression leads to:

$$\frac{1}{5} = \frac{|3\lambda - 8|}{\sqrt{35}\sqrt{\lambda^2 + 10}}$$

Squaring both sides to eliminate the square roots, we get $35(\lambda^2 + 10) = 25(3\lambda - 8)^2$. Expanding and simplifying this equation results in the quadratic:

$$19\lambda^2 - 120\lambda + 125 = 0$$

Factoring this quadratic yields $\lambda = 5$ and $\lambda = \frac{25}{19}$. Given the condition $\lambda_1 < \lambda_2$, we assign $\lambda_1 = \frac{25}{19}$ and $\lambda_2 = 5$.

### The Final Destination

We now locate the point $Q(38\lambda_1, 10\lambda_2, 2)$. Substituting our values for $\lambda_1$ and $\lambda_2$:

$$x = 38\left(\frac{25}{19}ight) = 50, \quad y = 10(5) = 50, \quad z = 2$$

Thus, the point is $Q(50, 50, 2)$. We calculate the perpendicular distance $d$ from $Q$ to the plane $P_1: 3x - 5y + z - 7 = 0$ using the standard distance formula:

$$d = \frac{|3(50) - 5(50) + 1(2) - 7|}{\sqrt{3^2 + (-5)^2 + 1^2}} = \frac{|150 - 250 + 2 - 7|}{\sqrt{35}} = \frac{105}{\sqrt{35}}$$

The problem asks for $d^2$:

$$d^2 = \frac{105^2}{35} = \frac{11025}{35} = 315$$

The final result of our spatial journey is **315**.

Enter Numerical Value:

Visualized Solution

Extracting Normal Vectors $n_{1}$ and $n_{2}$

Converting $sin θ$ to $cos θ$

Angle Between Two Planes

Evaluating the Expression

Solving for $λ$

Finding the Roots

Applying the Condition $λ_{1} < λ_{2}$

Coordinates of Point $Q$

Perpendicular Distance to Plane $P_{1}$

Evaluating Distance $d$

Square of the Distance

The Sigma Insight: Angle Between Two Planes

Analyzing the Setup

The Angle Trap

The Algebraic Battle

The Final Destination

Similar Questions

Let $P_{1} : 2 x + y - z = 3$ and $P_{2} : x + 2 y + z = 2$ be two planes. Then, which of the following statement(s) is (are) TRUE ?

The acute angle between the planes $P_{1}$ and $P_{2}$ , when $P_{1}$ and $P_{2}$ are the planes passing through the intersection of the planes $5 x + 8 y + 13 z - 29 = 0$ and $8 x - 7 y + z - 20 = 0$ and the points $(2, 1, 3)$ and $(0, 1, 2)$ , respectively, is

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

Let the acute angle bisector of the two planes $x - 2 y - 2 z + 1 = 0$ and $2 x - 3 y - 6 z + 1 = 0$ be the plane $P$ . Then which of the following points lies on $P$ ?

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

A plane which bisects the angle between the two given planes $2 x - y + 2 z - 4 = 0$ and $x + 2 y + 2 z - 2 = 0$ , passes through the point:

A tetrahedron has vertices $P (1, 2, 1)$ , $Q (2, 1, 3)$ , $R (- 1, 1, 2)$ and $O (0, 0, 0)$ . The angle between the faces $O P Q$ and $P QR$ is :

A tetrahedron has vertices at $O (0, 0, 0), A (1, 2, 1), B (2, 1, 3)$ and $C (- 1, 1, 2)$ . Then the angle between the faces OAB and ABC will be

Enter Numerical Value:

Visualized Solution

Extracting Normal Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n2​

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

Angle Between Two Planes

Evaluating the Expression

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

Finding the Roots

Applying the Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ1​<λ2​

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

Perpendicular Distance to Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​

Evaluating Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d