91Ó°ÊÓ

The cross product, also known as the vector product, is a fundamental operation in vector calculus used to compute a vector that is perpendicular to two given vectors. In our problem, we have vectors \( \vec{a} = \vec{i} + \vec{j} \) and \( \vec{b} = \vec{j} + 2\vec{k} \). The cross product is calculated using the formula:\[\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ 1 & 1 & 0 \ 0 & 1 & 2 \end{vmatrix}\]This involves computing the determinant of a 3x3 matrix, where the first row consists of the unit vectors \( \vec{i}, \vec{j}, \vec{k} \), and the second and third rows are the coefficients from vectors \( \vec{a} \) and \( \vec{b} \) respectively. Practically:

• \( \vec{i} \times \vec{j} = \vec{k} \)
• \( \vec{i} \times \vec{k} = -\vec{j} \)
• \( \vec{j} \times \vec{k} = \vec{i} \)

Using these properties, you can expand and simplify the expression to find the resultant vector, \( \vec{a} \times \vec{b} = \vec{k} - 2\vec{j} + 2\vec{i} \).

Calculating the magnitude is crucial when using the cross product to determine further values such as the area of a triangle. The magnitude of a vector \( \vec{v} = a\vec{i} + b\vec{j} + c\vec{k} \) is given by:\[||\vec{v}|| = \sqrt{a^2 + b^2 + c^2}\]In our case, the vector is \( 2\vec{i} - 2\vec{j} + \vec{k} \). The calculation proceeds as follows:\[||2\vec{i} - 2\vec{j} + \vec{k}|| = \sqrt{(2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3\]Breaking it down:

• Square each component of the vector: \( 2^2 = 4 \), \( (-2)^2 = 4 \), and \( 1^2 = 1 \).
• Add the squares together: 4 + 4 + 1 = 9.
• Take the square root of the sum: \( \sqrt{9} = 3 \).

The magnitude represents how long the vector is in space, which is essential when determining areas bounded by vectors.

The area of a triangle formed by two vectors can be calculated directly using the cross product. The formula is:\[\text{Area} = \frac{1}{2} \times ||\vec{a} \times \vec{b}||\]This formula tells us that the area is half the magnitude of the cross product of the vectors representing the triangle's sides. Applying it here, with \( ||\vec{a} \times \vec{b}|| = 3 \), the area becomes:\[\text{Area} = \frac{1}{2} \times 3 = \frac{3}{2}\]This provides the answer to our original question. Using these steps, we assured the statement is true: The area of the triangle with sides \( \vec{a} \) and \( \vec{b} \) is precisely \( \frac{3}{2} \), matching the provided area in the problem.

• Understand the link between the cross product and area calculation.
• The factor \( \frac{1}{2} \) accounts for the triangular shape, dividing the parallelogram formed by the vectors.
• Verifying calculations ensures accurate interpretation of geometric properties.