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34 A.C Find the equation of a plane which passes through the point (4, 3, 1) and is perpendicular to the line with direction ratios 2, – 4, 3. Also, find the distance of this plane from the origin. उस समतल का समीकरण ज्ञात करिए जो बिन्दु (4, 3, 1) से होकर गुजरता है तथा एक रेखा, जिसके दिक् अनुपात 2, – 4, 3 हैं, पर लम्ब है । इस समतल की मूल बिन्दु से दूरी भी ज्ञात करिए।​

34 AC Find the equation of a plane which passes through the point 4 3 1 and is perpendicular to the line with direction ratios 2 4 3 Also find the distance of t class=

Answer :

riteshkadam281213

Step-by-step explanation:

The correct option is B

2x + 3y - z = 17

Out of all the forms we have seen to write the equation of plane it is wise to use the point normal form here as we know the coordinates of a point and also the direction cosines of the perpendicular (normal). Equation of plane in point normal form is

a

(

x

−

x

1

)

+

b

(

y

−

y

1

)

+

c

(

z

−

z

1

)

=

0

Where a, b, c are the direction ratios of normal and

(

x

1

,

y

1

,

z

1

) are the coordinates of the point given.

So the equation will be -

2(x - 2) + 3(y -3) -1(z + 4) = 0

Or 2x + 3y - z = 17

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