Let OP $= ((α−1)/α) î + ĵ + k̂,$ OQ $= î + ((β−1)/β) ĵ + k̂$ and OR $= î + ĵ + (\frac{1}{2}) k̂$ be three vectors, where a $, b ∈ ℝ − \{0\}$ and O denotes the origin. If (OP × OQ) · OR $= 0$ and the point (α, β, 2) lies on the plane $3x + 3y − z + l = 0$ then the value of l is ______.

Answer: $5$

= ((α−1)/α) î + ĵ + k̂,

= î + ((β−1)/β) ĵ + k̂

and OR

= î + ĵ + (\frac{1}{2}) k̂

be three vectors, where a

, b ∈ ℝ − \{0\}

and O denotes the origin. If (OP × OQ) · OR

= 0

and the point (α, β, 2) lies on the plane

3x + 3y − z + l = 0

then the value of l is ______.

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Written solution

Let OP⃗ = (α-1)/αî + ĵ + k̂, OQ⃗ = î + (β-1)/βĵ + k̂ and OR⃗ = î + ĵ + 1/2k̂ be three vectors, where a, b ∈ R - {0} and O denotes the origin. If (OP⃗ × OQ⃗) · OR⃗ = 0 and the point (α, β, 2) lies on the plane 3x + 3y - z + l = 0 then the value of l is ______. $\vec{OP} = \dfrac{\alpha-1}{\alpha}\hat{i} + \hat{j} + \hat{k}$ $\vec{OQ} = \hat{i} + \dfrac{\beta-1}{\beta}\hat{j} + \hat{k}$ $\vec{OR} = \hat{i} + \hat{j} + \dfrac{1}{2}\hat{k}$ $(\vec{OP} \times \vec{OQ}) \cdot \vec{OR} = 0$ $(\alpha, \beta, 2) \in 3x + 3y - z + l = 0$ $\text{To find: } l$ $\text{Plan: } (1)\ (\vec{OP}\times\vec{OQ})\cdot\vec{OR}=0 \Rightarrow \det = 0$ $(2)\ \text{expand} \Rightarrow \text{a relation between } \alpha, \beta$ $(3)\ (\alpha,\beta,2) \Rightarrow l$ $\vec{OP} = \left(\dfrac{\alpha-1}{\alpha},\ 1,\ 1\right)$ $\vec{OQ} = \left(1,\ \dfrac{\beta-1}{\beta},\ 1\right)$ $\vec{OR} = \left(1,\ 1,\ \tfrac{1}{2}\right)$ $(\vec{OP}\times\vec{OQ})\cdot\vec{OR}=0 \Rightarrow \vec{OP},\ \vec{OQ},\ \vec{OR}\ \text{coplanar}$ $u = \dfrac{\alpha-1}{\alpha},\quad v = \dfrac{\beta-1}{\beta}$ $\vec{OP}=(u,1,1),\quad \vec{OQ}=(1,v,1),\quad \vec{OR}=\left(1,1,\tfrac{1}{2}\right)$ $\begin{vmatrix} u & 1 & 1 \\ 1 & v & 1 \\ 1 & 1 & \frac{1}{2} \end{vmatrix} = 0$ $= u\left(\dfrac{v}{2}-1\right) - \left(\dfrac{1}{2}-1\right) + (1-v)$ $= u\left(\dfrac{v}{2}-1\right) + \dfrac{3}{2} - v = 0$ $\dfrac{v}{2}-1 = -\dfrac{\beta+1}{2\beta},\qquad v-\dfrac{3}{2} = -\dfrac{\beta+2}{2\beta}$ $u(\beta+1) = \beta+2$ $\left(1-\dfrac{1}{\alpha}\right)(\beta+1) = \beta+2$ $(\beta+1) - \dfrac{\beta+1}{\alpha} = \beta+2$ $\alpha = -(\beta+1) \Rightarrow \alpha + \beta = -1$ $3\alpha + 3\beta - 2 + l = 0$ $3(\alpha + \beta) - 2 + l = 0$ $3(-1) - 2 + l = 0$ $-5 + l = 0 \Rightarrow l = 5$ $A(-2,1,2):\ 3(-2)+3(1)-2+5 = 0\ \checkmark$ $l = 5$

Written solution

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