91Ó°ÊÓ

When dealing with planes in three-dimensional space, a normal vector is essential. This vector is perpendicular to the plane and helps define its orientation. To find the normal vector of a plane defined by an equation such as \[ ax + by + cz = d \]we look at the coefficients of the variables. Therefore, for the plane equation \( x + y = 1 \), the normal vector \( \mathbf{n}_1 \) is derived as \((1, 1, 0)\). This means it points in such a manner that is perpendicular to the x and y axes only since there's no \( z \) component.
Similarly, for the plane \( 2x + y - 2z = 2 \), the normal vector \( \mathbf{n}_2 \) is \((2, 1, -2)\). This vector accounts for all x, y, and z components present in the equation.

The dot product, also known as the scalar product, is an operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation combines vectors and provides important information like the angle between them. For vectors \( \mathbf{a} \) and \( \mathbf{b} \), the dot product is calculated as:\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]In our exercise, the dot product of \( \mathbf{n}_1 = (1, 1, 0) \) and \( \mathbf{n}_2 = (2, 1, -2) \) is calculated by multiplying corresponding components and summing them up:- \( 1 \times 2 \)- \( 1 \times 1 \)- \( 0 \times (-2) \)Adding these gives a total of 3.

The magnitude of a vector provides the length of the vector and is essential in angle calculations. For a vector \( \mathbf{a} = (a_1, a_2, a_3) \), its magnitude is calculated using the formula:\[ \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]In the exercise, for normal vector \( \mathbf{n}_1 = (1, 1, 0) \), the magnitude is:\[ \| \mathbf{n}_1 \| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2} \]For \( \mathbf{n}_2 = (2, 1, -2) \), the magnitude is:\[ \| \mathbf{n}_2 \| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3 \]

The cosine formula provides a convenient way to calculate the angle between two vectors. It relies on the dot product and magnitudes of the vectors. Given two vectors \( \mathbf{a} \) and \( \mathbf{b} \), the formula is:\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \| \| \mathbf{b} \|} \]In the example from the exercise, we have calculated:- Dot product: 3- Magnitudes: \( \| \mathbf{n}_1 \| = \sqrt{2} \), \( \| \mathbf{n}_2 \| = 3 \)Plugging these into the formula gives:\[ \cos \theta = \frac{3}{\sqrt{2} \cdot 3} = \frac{1}{\sqrt{2}} \]The inverse cosine of \( \frac{1}{\sqrt{2}} \) is 45° or \( \frac{\pi}{4} \) radians, indicating a right angle between the original planes.