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CBSE Class 12-science Answered

Solve the following question:
Asked by joykphukan | 23 Dec, 2015, 06:26: PM
answered-by-expert Expert Answer
bold Let bold space bold the bold space bold equation bold space bold of bold space bold the bold space bold plane bold space bold passing bold space bold through bold space bold the bold space bold line bold space bold of bold space bold intersection bold space bold of bold space bold r with bold rightwards arrow on top bold. open parentheses bold 2 bold i with bold overbrace on top bold plus bold 6 bold j with bold overbrace on top close parentheses bold plus bold 12 bold plus bold lambda open square brackets bold r with bold rightwards arrow on top bold. open parentheses bold 3 bold i with bold overbrace on top bold minus bold j with bold overbrace on top bold plus bold 4 bold k with bold overbrace on top close parentheses close square brackets bold equals bold 0 bold rightwards double arrow bold r with bold rightwards arrow on top bold. open square brackets open parentheses bold 2 bold plus bold 3 bold lambda close parentheses bold i with bold hat on top bold plus open parentheses bold 6 bold minus bold lambda close parentheses bold j with bold hat on top bold plus bold 4 bold lambda bold k with bold hat on top close square brackets bold plus bold 12 bold equals bold 0 bold. bold. bold. bold left parenthesis bold 1 bold right parenthesis bold Writing bold space bold in bold space bold normal bold space bold form fraction numerator bold r with bold rightwards arrow on top bold. open square brackets open parentheses bold 2 bold plus bold 3 bold lambda close parentheses bold i with bold hat on top bold plus open parentheses bold 6 bold minus bold lambda close parentheses bold j with bold hat on top bold plus bold 4 bold lambda bold k with bold hat on top close square brackets over denominator square root of open parentheses bold 2 bold plus bold 3 bold lambda close parentheses to the power of bold 2 bold plus open parentheses bold 6 bold minus bold lambda close parentheses to the power of bold 2 bold plus open parentheses bold 4 bold lambda close parentheses to the power of bold 2 end root end fraction bold equals fraction numerator bold minus bold 12 over denominator square root of open parentheses bold 2 bold plus bold 3 bold lambda close parentheses to the power of bold 2 bold plus open parentheses bold 6 bold minus bold lambda close parentheses to the power of bold 2 bold plus open parentheses bold 4 bold lambda close parentheses to the power of bold 2 end root end fraction bold Distance bold space bold from bold space bold the bold space bold origin bold space bold equals open vertical bar fraction numerator bold minus bold 12 over denominator square root of open parentheses bold 2 bold plus bold 3 bold lambda close parentheses to the power of bold 2 bold plus open parentheses bold 6 bold minus bold lambda close parentheses to the power of bold 2 bold plus open parentheses bold 4 bold lambda close parentheses to the power of bold 2 end root end fraction close vertical bar open vertical bar fraction numerator bold minus bold 12 over denominator square root of open parentheses bold 2 bold plus bold 3 bold lambda close parentheses to the power of bold 2 bold plus open parentheses bold 6 bold minus bold lambda close parentheses to the power of bold 2 bold plus open parentheses bold 4 bold lambda close parentheses to the power of bold 2 end root end fraction close vertical bar bold equals bold 1 bold rightwards double arrow open parentheses bold 2 bold plus bold 3 bold lambda close parentheses to the power of bold 2 bold plus open parentheses bold 6 bold minus bold lambda close parentheses to the power of bold 2 bold plus open parentheses bold 4 bold lambda close parentheses to the power of bold 2 bold equals bold 144 bold rightwards double arrow bold 26 bold lambda to the power of bold 2 bold plus bold 40 bold equals bold 144 bold rightwards double arrow bold lambda to the power of bold 2 bold equals bold 4 bold rightwards double arrow bold lambda bold equals bold plus-or-minus bold 2 bold Using bold space bold the bold space bold value bold space bold of bold space bold lambda bold space bold in bold space bold left parenthesis bold 1 bold right parenthesis bold comma bold space bold we bold space bold get bold space bold the bold space bold equation bold space bold of bold space bold plane bold space bold as bold r with bold rightwards arrow on top bold. open square brackets open parentheses bold 2 bold plus-or-minus bold 6 close parentheses bold i with bold hat on top bold plus open parentheses bold 6 bold minus-or-plus bold 2 close parentheses bold j with bold hat on top bold plus-or-minus bold 8 bold k with bold hat on top close square brackets bold plus bold 12 bold equals bold 0 bold rightwards double arrow bold r with bold rightwards arrow on top bold. open parentheses bold 8 bold i with bold overbrace on top bold minus bold 4 bold j with bold overbrace on top bold plus bold 8 bold k with bold overbrace on top close parentheses bold plus bold 12 bold equals bold 0 bold space bold space bold or bold space bold r with bold rightwards arrow on top bold. open parentheses bold minus bold 4 bold i with bold hat on top bold plus bold 8 bold j with bold overbrace on top bold minus bold 8 bold k with bold overbrace on top close parentheses bold plus bold 12 bold equals bold 0 bold rightwards double arrow bold r with bold rightwards arrow on top bold. open parentheses bold 2 bold i with bold overbrace on top bold minus bold j with bold overbrace on top bold plus bold 2 bold k with bold overbrace on top close parentheses bold plus bold 3 bold equals bold 0 bold space bold space bold or bold space bold r with bold rightwards arrow on top bold. open parentheses bold minus bold i with bold hat on top bold plus bold 2 bold j with bold overbrace on top bold minus bold 2 bold k with bold overbrace on top close parentheses bold plus bold 3 bold equals bold 0 bold In bold space bold cartesian bold space bold form bold space bold 2 bold x bold minus bold y bold plus bold 2 bold z bold plus bold 3 bold equals bold 0 bold space bold or bold space bold minus bold x bold plus bold 2 bold y bold minus bold 2 bold z bold plus bold 3 bold equals bold 0
Answered by satyajit samal | 24 Dec, 2015, 04:10: PM

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CBSE 12-science - Maths
Q 48
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Asked by lekhakarthikeyan | 16 Jul, 2018, 03:58: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Maths
If  is the normal from the origin to the plane, and  is the unit vector along . P(x, y, z) be any point on the plane and  is perpendicular to . Find the equation of plane in the normal form.
Asked by Topperlearning User | 26 Feb, 2015, 10:10: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Maths
Let  be the vector normal to the plane and  be the position vector of the point through which the plane passes. Then find the equation of plane.
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Maths
Find the equation of the plane, which is at a distance of 5 unit from the origin and has as a normal vector.
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
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CBSE 12-science - Maths
Find the equation of the plane whose distance from the origin is 8 units and the direction ratios of the normal are 6, – 3, – 2.
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
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CBSE 12-science - Maths
Find the equation of the plane is . Find the perpendicular distance of the plane from the origin.
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
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CBSE 12-science - Maths
Write the normal and Cartesian form of the plane .
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
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CBSE 12-science - Maths
Find the equation of the plane passing through the points (–1, 4, – 3), (3, 2, – 5) and (– 3, 8, – 5).
Asked by Topperlearning User | 26 Feb, 2015, 10:50: AM
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CBSE 12-science - Maths
Find the equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points (3, 4, – 1) and (2, – 1, 5).
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
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CBSE 12-science - Maths
Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane 2x – 3y + 6z + 7 = 0.
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
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