Question

The angle between \(\hat{i}\) and the line of intersection of the plane \(\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})=0\) and \(\vec{r} \cdot(3 \hat{i}+3 \hat{j}+\hat{k})=0\) is (1) \(\cos ^{-1}\left(\frac{1}{3}\right)\) (2) \(\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)\) (3) \(\cos ^{-1}\left(\frac{2}{\sqrt{3}}\right)\) (4) \(\cos ^{-1}\left(\frac{7}{\sqrt{122}}\right)\)

Short answer

\(\cos^{-1}\left(\frac{7}{\sqrt{122}}\right)\)

Step-by-step solution

Understand the Problem

The problem involves finding the angle between the vector \(\text{\hat{i}}\) and the line of intersection of two planes given by their normal vectors.

Identify the Normal Vectors

The normal vector of the first plane \(\text{\vec{r} \cdot(\hat{i}+2\hat{j}+3\hat{k})=0}\) is \(\text{\hat{i}+2\hat{j}+3\hat{k}}\). The normal vector of the second plane \(\text{\vec{r} \cdot(3\hat{i}+3\hat{j}+\hat{k})=0}\) is \(\text{3\hat{i}+3\hat{j}+\hat{k}}\).

Find the Cross Product

The direction vector of the line of intersection is given by the cross product of the two normal vectors. \[ \text{\vec{n_1}} = \hat{i} + 2\hat{j} + 3\hat{k} \text {and} \text{\vec{n_2}} = 3\hat{i} + 3\hat{j} + \hat{k} \] \[ \text {\vec{d} = \vec{n_1} \times \vec{n_2}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 3 & 3 & 1 \end{vmatrix} \] = \[ (2 \cdot 1 - 3 \cdot 3)\hat{i} - (1 \cdot 1 - 3 \cdot 3)\hat{j} + (1 \cdot 3 - 2 \cdot 3)\hat{k} = (-7)\hat{i} + 8\hat{j} + (-3)\hat{k}\]

Calculate Magnitude of Direction Vector

Find the magnitude of the direction vector \(\text{\vec{d}}\). \[ \|\vec{d}\| = \sqrt{(-7)^2 + 8^2 + (-3)^2} = \sqrt{49 + 64 + 9} = \sqrt{122} \]

Find the Dot Product with \(\text{\hat{i}}\)

Calculate the dot product of \(\text{\vec{d}}\) with \(\text{\hat{i}}\). \[ \vec{d} \cdot \hat{i} = (-7) \cdot 1 + (8) \cdot 0 + (-3) \cdot 0 = -7 \]

Determine Cosine of the Angle

Use the dot product formula to find the cosine of the angle \(\theta\). \[ \cos \theta = \frac{\vec{d} \cdot \hat{i}}{\|\vec{d}\| \cdot \|\hat{i}\|} = \frac{-7}{\sqrt{122} \cdot 1} = -\frac{7}{\sqrt{122}} \] The angle \(\theta\) is then \[ \theta = \cos^{-1}\left(\frac{7}{\sqrt{122}}\right) \]

Key concepts

Normal Vectors

In three-dimensional space, a plane can be defined using a point and a normal vector. The normal vector is perpendicular to the plane's surface.
For the problem at hand, we have two planes defined by their normal vectors. The first plane, \(\text{\vec{r} \cdot(\text{\hat{i}}+2\text{\hat{j}}+3\text{\hat{k}})=0}\), has a normal vector of \(\hat{i}+2\hat{j}+3\hat{k}\). The second plane, \(\text{\vec{r} \cdot(3\text{\hat{i}}+3\hat{j}+\text{\hat{k}})=0}\), has a normal vector of \(3\hat{i}+3\hat{j}+\hat{k}\).
These normal vectors help us in defining the orientation of the planes and are crucial for our calculations.

Cross Product

The cross product of two vectors results in a third vector that is perpendicular to both of the original vectors.
To find the direction vector of the line of intersection between the two planes, we calculate the cross product of their normal vectors: \(\vec{n_1} = \hat{i} + 2\hat{j} + 3\hat{k}\) and \(\vec{n_2} = 3\hat{i} + 3\hat{j} + \hat{k}\). Using the determinant method, we get:
\[ \vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 3 & 3 & 1 \end{vmatrix} = (-7)\hat{i} + 8\hat{j} + (-3)\hat{k} \]
This cross product gives us the direction vector \(\vec{d}\), indicating the line's orientation where the two planes intersect.

Dot Product

The dot product of two vectors is a scalar value that measures the magnitude of one vector in the direction of another.
To find the angle between the vector \(\hat{i}\) and our direction vector \(\vec{d} = (-7)\hat{i} + 8\hat{j} + (-3)\hat{k}\), we first compute the dot product:
\[ \vec{d} \cdot \hat{i} = (-7) \cdot 1 + 8 \cdot 0 + (-3) \cdot 0 = -7 \]
This result is necessary for determining the cosine of the angle between the two vectors.

Angle Calculation

Using the dot product, we can find the cosine of the angle \(\theta\) between two vectors by dividing their dot product by the product of their magnitudes.
First, we find the magnitude of the direction vector \(\vec{d}\) as \ \[ \|\vec{d}\| = \sqrt{(-7)^2 + 8^2 + (-3)^2} = \sqrt{122} \]
Then, using the dot product formula: \[ \cos \theta = \frac{\vec{d} \cdot \hat{i}}{\|\vec{d}\| \|\text{\hat{i}}\|} = \frac{-7}{\sqrt{122} \cdot 1} = -\frac{7}{\/\sqrt{122}} \]
This calculation gives us: \( \theta = \cos^{-1}\left(\frac{7}{\/\sqrt{122}}\right) \).

Vector Magnitude

The magnitude of a vector represents its length. It is calculated by taking the square root of the sum of the squares of its components.
For the direction vector \(\vec{d} = (-7)\hat{i} + 8\hat{j} + (-3)\hat{k}\), the magnitude \|\vec{d}\|\ is given by: \[ \| \vec{d} \| = \sqrt{(-7)^2 + 8^2 + (-3)^2} = \sqrt{49 + 64 + 9} = \sqrt{122} \]
This value is essential for calculating the cosine of the angle between the direction vector and \(\hat{i}\).