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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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$

Similar Questions

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and $\lambda$ is a real number,then for what value of $\lambda$ are the vectors $\vec{a} + 2\vec{b} + 3\vec{c}$,$\lambda\vec{b} + 4\vec{c}$,and $(2\lambda - 1)\vec{c}$ non-coplanar?

The volume of the parallelepiped whose conterminous edges are $\vec{a} = \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = 2\hat{i} - 4\hat{j} + 5\hat{k}$,and $\vec{c} = 3\hat{i} - 5\hat{j} + 2\hat{k}$ is:

If $a = i + j + k$,$b = 4i + 3j + 4k$,and $c = i + \alpha j + \beta k$ are coplanar vectors and $|c| = \sqrt{3}$,then:

If $3 \hat{i}-2 \hat{j}-\hat{k}$,$2 \hat{i}+3 \hat{j}-4 \hat{k}$,$-\hat{i}+\hat{j}+2 \hat{k}$ and $4 \hat{i}+5 \hat{j}+\lambda \hat{k}$ are respectively the position vectors of four coplanar points $P, Q, R$ and $S$,then $\lambda=$

The volume of a parallelepiped whose coterminous edges are represented by the vectors $\overrightarrow{OA} = (2, 1, 1)$,$\overrightarrow{OB} = (3, -1, 1)$,and $\overrightarrow{OC} = (-1, 1, -1)$ is . . . . . . cubic units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module