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in Three-dimensional geometry by (15 points)
edited by

Find the equntion of the plane through the line of intersection of \( \vec{r} \cdot(2 \hat{i}-3 \hat{j}+4 \hat{k})=1 \) and \( \vec{r} \cdot(\hat{i}-\hat{j})+4=0 \) and perpendicular to the plane \( \vec{r} \cdot(2 \vec{i}-\vec{j}+\hat{h})+8=0 \). Hence, find whether the plane thus obtained contains the line \( x-1=2 y-4=3 z-12 \).

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2 Answers

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by (15 points)
image
This question uses 2 formula
First: You have to use the formula of equations of plane passing through the intersection of two given planes.
Second: You have to use the formula of Angle between two planes.
You have to convert every equations of planes in Cartesian Form.
+1 vote
by (35.7k points)
edited by

Given plane is \(\vec r. (2\hat i-\hat j + \hat k) + 8 = 0 \)........(1)

Required plane passes through the line of intersection of passes \(\vec r. (2\hat i-3\hat j + 4\hat k) = 1\, and \,\vec r. (\hat i-\hat j) + 4 = 0\)

∴ Equation of require plane is \(\vec r . (2\hat i-3\hat j + 4\hat k) + λ(\hat i - \hat j) + 1 + 4λ = 0\)

\(\vec r. ((2 + λ)\hat i + (-3- λ)\hat j + 4\hat k) + 1 + 4 λ = 0 ....(2) \)

∴ Plane (1) & (2) perpendicular to each other 

∴ 2 (2 + λ) + -1 x - (3 + λ) + 1 x 4 = 0

= 4 + 2λ + 3 + λ + 4 = 0

= 3λ + 11 = 0

= λ = -11/3

∴ Equation of required plane is \(\vec r. ((2-\frac {11}{3})\hat i + (-3 + \frac {11}{3}) \hat j + 4\hat k) + 1 + 4 x - \frac {11}{3} = 0\)

\(\vec r.(-5 \hat i + 2\hat j + 12 \hat k) + 3-44 = 0\)

\(\vec r.(-5\hat i + 2\hat j + 12 \hat k) - 41 = 0 ......(3)\)

∴ Direction ratios of normal to plane (3) are -5, 2 &12.

Given line is x-1 = 2y-4 = 3 z -12

\(\frac {x-1}{1} = \frac {y-2}{\frac 12} = \frac {z-4}{\frac 13}\)

∴ Direction ratios of line are 1, 1/2 & 1/3 and it passes through point (1, 2, 4)

∴  -5 x 1 + 2 x 1/2 + 12 x 1/3 = -5 + 1+ 4

= -5 + 5 =0

Hence, line is perpendicular to normal of the plane (3).

∴ Line is paralleled to plane (3)

Also \((\hat i + 2\hat j + 4\hat k).(-5 \hat + 2\hat j + 12\hat k) -41 = 0\) (from (3) )

= -5 + 4 + 48 - 41 = 0

= 6 = 0 (Not satisfied)

Hence, point (1, 2, 4) does not lie on plane (3).

Therefore, given line is paralleled to plane (3) but plane did not contains the line.

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