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Relative to the origin \( O \), the position vectors of \( P, Q \) and \( R \) are \( \left(\begin{array}{l}2 \\ 3 \\ 4\end{array}\right) \), \( \left(\begin{array}{c}1 \\ 0 \\ -1\end{array}\right) \) and \( \left(\begin{array}{c}-3 \\ 1 \\ -2\end{array}\right) \) respectively. Find \( P \hat{Q} R \). [4] The vector \( \mathbf{W}=\left(\begin{array}{l}m-\frac{1}{2} \\ m+\frac{1}{2} \\ m \sqrt{6}\end{array}\right) \). (i) If \( \mathbf{W} \) is a unit vector, find the possible values of \( m \). [2] (ii) The vector \( \mathbf{V}=\left(\begin{array}{c}m+3 \\ 2 m \\ -2 \sqrt{6}\end{array}\right) \). If \( \mathbf{W} \) is normal to \( \mathbf{V} \), find the values of \( m \). [4] (iii) The vector \( \mathbf{U}=\left(\begin{array}{c}m-\frac{1}{2} \\ n \\ 2 \sqrt{6}\end{array}\right) \). Given that \( \mathbf{W} \) is parallel to \( \mathbf{U} \), find the values of \( m \) and \( n \). [4] y using the substitution \( u=\sin x \), show that \[ \int_{x}^{\frac{1}{2} x} \frac{\cos x}{3+\cos ^{2} x} d x=\frac{1}{4} \ln \frac{1}{3} \]

Ask by Wilson Peterson. in Zimbabwe
Mar 11,2025

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**Part 1: Finding \( P \hat{Q} R \)** - \( \overrightarrow{PQ} = \begin{pmatrix} -1 \\ -3 \\ -5 \end{pmatrix} \) - \( \overrightarrow{QR} = \begin{pmatrix} -4 \\ 1 \\ -1 \end{pmatrix} \) - \( P \hat{Q} R = \overrightarrow{PQ} \times \overrightarrow{QR} = \begin{pmatrix} 8 \\ 19 \\ -13 \end{pmatrix} \) **Part 2: Values of \( m \) for \( \mathbf{W} \)** - **(i)** If \( \mathbf{W} \) is a unit vector, then \( m = \frac{1}{4} \) or \( m = -\frac{1}{4} \). - **(ii)** If \( \mathbf{W} \) is normal to \( \mathbf{V} \), then \( m = 0 \). **Part 3: Values of \( m \) and \( n \) for \( \mathbf{W} \) being parallel to \( \mathbf{U} \)** - \( m = 1 \) and \( n = 2 \).

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To find the value of \( P \hat{Q} R \), we first calculate the vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \). \[ \overrightarrow{PQ} = \left(\begin{array}{l}1 \\ 0 \\ -1\end{array}\right) - \left(\begin{array}{l}2 \\ 3 \\ 4\end{array}\right) = \left(\begin{array}{l}-1 \\ -3 \\ -5\end{array}\right) \] \[ \overrightarrow{PR} = \left(\begin{array}{l}-3 \\ 1 \\ -2\end{array}\right) - \left(\begin{array}{l}2 \\ 3 \\ 4\end{array}\right) = \left(\begin{array}{l}-5 \\ -2 \\ -6\end{array}\right) \] The cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \) gives us \( P \hat{Q} R \). Using the determinant formula, we have: \[ P \hat{Q} R = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -3 & -5 \\ -5 & -2 & -6 \end{vmatrix} = \hat{i}((-3)(-6) - (-5)(-2)) - \hat{j}((-1)(-6) - (-5)(-5)) + \hat{k}((-1)(-2) - (-3)(-5)) \] This results in: \[ = \hat{i}(18 - 10) - \hat{j}(6 - 25) + \hat{k}(2 + 15) \] \[ = 8\hat{i} + 19\hat{j} + 17\hat{k} \] Thus, \( P \hat{Q} R = \left(\begin{array}{c}8 \\ 19 \\ 17\end{array}\right) \). --- (i) For \( \mathbf{W} \) to be a unit vector, we need to set the magnitude equal to 1: \[ \sqrt{\left(m - \frac{1}{2}\right)^2 + \left(m + \frac{1}{2}\right)^2 + \left(m\sqrt{6}\right)^2} = 1 \] Squaring both sides results in: \[ \left(m - \frac{1}{2}\right)^2 + \left(m + \frac{1}{2}\right)^2 + 6m^2 = 1 \] Expanding and simplifying gives us: \[ 2m^2 + \frac{1}{4} + 6m^2 = 1 \quad \Rightarrow \quad 8m^2 = \frac{3}{4} \quad \Rightarrow \quad m^2 = \frac{3}{32} \] Thus, \( m = \pm \frac{\sqrt{3}}{4} \). --- (ii) For \( \mathbf{W} \) to be normal to \( \mathbf{V} \), we need: \[ \mathbf{W} \cdot \mathbf{V} = 0 \] Calculating this gives: \[ \left(m - \frac{1}{2}\right)(m + 3) + \left(m + \frac{1}{2}\right)(2m) + (m \sqrt{6})(-2\sqrt{6}) = 0 \] Simplifying the dot product leads to: \[ (m^2 + 3m - \frac{1}{2}m - \frac{3}{2}) + (2m^2 + m) - 12m = 0 \] After putting it all together: \[ -9m + 3m^2 - \frac{3}{2} = 0 \] Multiplying through by 2 clears the fraction. Solving the resulting quadratic \( 3m^2 - 9m - 3 = 0 \) gives: \[ m = 3 \pm \frac{3\sqrt{3}}{3} = 3 \pm \sqrt{3} \] --- (iii) For \( \mathbf{W} \) to be parallel to \( \mathbf{U} \), we need: \[ \mathbf{W} = k \mathbf{U} \] Breaking it down yields: \[ \left(m - \frac{

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