Let overrightarrow{OP} = frac{alpha-1}{alpha} | vedclass

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(A) Given the scalar triple product $(\overrightarrow{OP} 	imes \overrightarrow{OQ}) \cdot \overrightarrow{OR} = 0$,the determinant of the components

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Given the scalar triple product $(\overrightarrow{OP} 	imes \overrightarrow{OQ}) \cdot \overrightarrow{OR} = 0$,the determinant of the components must be zero:$\begin{vmatrix} \frac{\alpha-1}{\alpha} & 1 & 1 \ 1 & \frac{\beta-1}{\beta} & 1 \ 1 & 1 & \frac{1}{2} \end{vmatrix} = 0$Expanding the determinant:$\frac{\alpha-1}{\alpha} (\frac{\beta-1}{2\beta} - 1) - 1 (\frac{1}{2} - 1) + 1 (1 - \frac{\beta-1}{\beta}) = 0$$\frac{\alpha-1}{\alpha} (\frac{\beta-1-2\beta}{2\beta}) - 1 (-\frac{1}{2}) + 1 (\frac{\beta-\beta+1}{\beta}) = 0$$\frac{\alpha-1}{\alpha} (\frac{-\beta-1}{2\beta}) + \frac{1}{2} + \frac{1}{\beta} = 0$Multiplying by $2\alpha\beta$: $-(\alpha-1)(\beta+1) + \alpha\beta + 2\alpha = 0$$-(\alpha\beta + \alpha - \beta - 1) + \alpha\beta + 2\alpha = 0$$-\alpha\beta - \alpha + \beta + 1 + \alpha\beta + 2\alpha = 0$$\alpha + \beta + 1 = 0 \Rightarrow \alpha + \beta = -1$  $(1)$Since the point $(\alpha, \beta, 2)$ lies on the plane $3x + 3y - z + l = 0$:$3\alpha + 3\beta - 2 + l = 0$$3(\alpha + \beta) - 2 + l = 0$Substituting $(1)$ into the equation:$3(-1) - 2 + l = 0$$-3 - 2 + l = 0 \Rightarrow l = 5$

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