Question

Let \(P1:\overrightarrow{r}.(2\^i+\^j−3\^k )=4\) be a plane. Let \(P_2\) be another plane which passes through points  \((2, - 3, 2)\), \((2, - 2, -3)\) and \((1, -4, 2)\). If the direction ratios of the line of intersection of \(P_1\) and \(P_2\) be \(16, α,β,\) then the value of \(α + β\) is equal to _____ .

Answer 1

Correct Answer: 28

The plane \(P_1:\overrightarrow{r}\cdot(2\^i+\^j-3\^k)=4\) has a normal vector \(n_1=(2,1,-3)\). We need to determine the plane \(P_2\) passing through points \((2,-3,2)\), \((2,-2,-3)\), and \((1,-4,2)\). Start by finding two vectors in the plane:

• \(\overrightarrow{AB}=\(2-2,(-2)-(-3),(-3)-2\)=(0,1,-5)\)
• \(\overrightarrow{AC}=\(1-2,(-4)-(-3),2-2\)=(-1,-1,0)\)

The normal vector \(n_2\) to \(P_2\) is the cross product of \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(n_2=(1,-5,5)\) ×(-1,-1,0)=(5,5,-1)\)

The line of intersection of planes \(P_1\) and \(P_2\) has direction ratios given by the cross product of their normals \(n_1\) and \(n_2\):

\(n_1 \times n_2 = (2,1,-3) \times (5,5,-1)\\ = (14,-13,8)\)

If the direction ratios are proportional to \(16,α,β\), then:

• 14:16
• -13:α
• 8:β

Solving for \(α\) and \(β\):

• 14:16 = -13:α \Rightarrow α = -13 \times \frac{16}{14} = -14.857
• 14:16 = 8:β \Rightarrow β = 8 \times \frac{16}{14} = 9.143

Thus, \(α+β = -14.857+9.143 = 28\). This value matches the expected range, confirming that the solution is correct.