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GCD(1260,1680,1920) - go-checkin.com

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 #### 8Question: Given three vertices of a cube at \((1, 2, 3)\), \((1, 5, 3)\), and \((4, 2, 3)\), find the coordinates of the fourth vertex that is diagonally opposite to \((1, 2, 3)\) on the same face, assuming all coordinates are integers.

📌 Solution: The given points lie on the plane \(z = 3\). The vector from \((1, 2, 3)\) to \((1, 5, 3)\) is \((0, 3, 0)\), and the vector to \((4, 2, 3)\) is \((3, 0, 0)\). The fourth vertex on the same face is obtained by adding these vectors: \((1 + 3, 2 + 3, 3) = (4, 5, 3)\).

📌 \boxed{(4, 5, 3)}

📌 Question: Find the value of \(x\) such that the vectors \(\begin{pmatrix} x \\ 2 \\ -1 \end{pmatrix}\) and \(\begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}\) are orthogonal.

📌 Solution: Two vectors are orthogonal if their dot product is zero. Compute the dot product: \(3x + (2)(-1) + (-1)(4) = 3x - 2 - 4 = 3x - 6\). Set this equal to zero: \(3x - 6 = 0 \implies x = 2\).

📌 Question: Find the vector \(\mathbf{v} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}\) such that \(\mathbf{v} \times \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 5 \\ 7 \\ -4 \end{pmatrix}\).

📌 Solution: The cross product \(\mathbf{v} \times \mathbf{a}\) is computed as:

📌 \begin{pmatrix} b \cdot 3 - c \cdot (-1) \\ c \cdot 2 - a \cdot 3 \\ a \cdot (-1) - b \cdot 2 \end{pmatrix} = \begin{pmatrix} 3b + c \\ 2c - 3a \\ -a - 2b \end{pmatrix}.

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