The problem asks us to find a vector $\vec{p}$ that is orthogonal to two given vectors $\vec{a} = (1; 4; 2)$ and $\vec{b} = (2; -1; 3)$, with the condition that the magnitude of $\vec{p}$ is $|\vec{p}| = 4\sqrt{278}$, and that $\vec{p}$ forms an acute angle with the $OY$-axis.

Steps:

1. Orthogonal Vector Condition: To find $\vec{p}$, we use the fact that it must be orthogonal to both $\vec{a}$ and $\vec{b}$. The cross product $\vec{p} = \vec{a} \times \vec{b}$ gives a vector that is perpendicular to both.

2. Compute Cross Product: The cross product of two vectors in 3D is computed as follows: $\vec{p} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 2 & -1 & 3 \end{vmatrix}$

   Expanding the determinant: $\vec{p} = \hat{i} \left( 4 \times 3 - 2 \times (-1) \right) - \hat{j} \left( 1 \times 3 - 2 \times 2 \right) + \hat{k} \left( 1 \times (-1) - 4 \times 2 \right)$ $\vec{p} = \hat{i} (12 + 2) - \hat{j} (3 - 4) + \hat{k} (-1 - 8)$ $\vec{p} = 14\hat{i} + \hat{j} - 9\hat{k}$ So, $\vec{p} = (14, 1, -9)$.

3. Scaling to Satisfy the Magnitude Condition: The magnitude of this vector is: $|\vec{p}| = \sqrt{14^2 + 1^2 + (-9)^2} = \sqrt{196 + 1 + 81} = \sqrt{278}$

   The desired magnitude is $4\sqrt{278}$, so we need to scale $\vec{p}$ by 4: $\vec{p}_{\text{final}} = 4 \cdot (14, 1, -9) = (56, 4, -36)$

4. Ensuring Acute Angle with the OY-axis: The $OY$-axis is represented by the vector $\vec{j} = (0, 1, 0)$. The dot product $\vec{p} \cdot \vec{j}$ gives us the cosine of the angle between $\vec{p}$ and the $OY$-axis. Since the second component of $\vec{p}$ is positive (4), the angle is acute, as required.

Thus, the vector $\vec{p}$ that satisfies all the conditions is $\vec{p} = (56, 4, -36)$.

Do you want any further details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How is the cross product used to find orthogonal vectors in 3D space?
2. What properties of a vector ensure it forms an acute angle with a specific axis?
3. How do you compute the magnitude of a vector in 3D space?
4. How can scaling a vector change its magnitude without changing its direction?
5. Why is the dot product useful for determining the angle between two vectors?

Tip: When working with vectors in 3D, the cross product is a very useful tool for finding orthogonal vectors, especially in physics and engineering applications.