Find the vector equation in scalar product form of the plane tha

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171. Find the equation of the plane which is parallel to the x-axis and has intercepts 5 and 7 on the y and z-axis, respectively.
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172. Find the D.C.’s of the perpendicular from origin to the plane  straight r with rightwards arrow on top. space open parentheses negative 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 6 space straight k with hat on top close parentheses plus 14 space equals space 0.  Find also the distance of the plane from the origin. 
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176. Find the vector equation of the following planes in scalar product form:
straight r with rightwards arrow on top space equals space straight i with hat on top space minus space straight j with hat on top space plus space straight lambda open parentheses straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top close parentheses space plus space straight mu space open parentheses straight i with hat on top space minus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses
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177. Find the vector equation of the following planes in scalar product form:
straight r with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space straight k with hat on top space plus space straight lambda space straight i with hat on top space plus space straight mu space open parentheses straight i with hat on top space minus space 2 space space straight j with hat on top space minus space straight k with hat on top close parentheses

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178. Find the vector equation of the plane in scalar product form
straight r with rightwards arrow on top space equals space straight i with hat on top space minus space straight j with hat on top space plus space straight lambda space open parentheses straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top close parentheses space plus space straight mu space open parentheses 4 straight i with hat on top space minus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses.

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179.

Find the vector equation in scalar product form of the plane that contains the lines.
                    straight r with rightwards arrow on top space equals space left parenthesis straight i with hat on top space plus space straight j with hat on top right parenthesis space plus space straight s space left parenthesis straight i with hat on top space plus 2 space straight j with hat on top space minus space straight k with hat on top right parenthesis
and               straight r with rightwards arrow on top space equals space open parentheses straight i with hat on top plus straight j with hat on top close parentheses plus straight t open parentheses negative straight i with hat on top space plus space straight j with hat on top space minus space 2 space straight k with hat on top close parentheses


The equation of lines are
                    straight r with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space plus space straight s space open parentheses straight i with hat on top space plus space 2 space straight j with hat on top space minus space straight k with hat on top close parentheses                                 ...(1)
and               straight r with rightwards arrow on top space equals space open parentheses straight i with hat on top plus straight j with hat on top close parentheses plus straight t open parentheses negative straight i with hat on top space plus space straight j with hat on top space minus space 2 space straight k with hat on top close parentheses                                ...(2)
Both the lines (1) and (2) pass through the point (1, 1, 0) whose position vector is straight i with hat on top space plus space straight j with hat on top.
Hence both the lines are intersecting and are coplanar. 
The equation of any plane containing the line (1) is
                        A (x - 1) + B (y - 1) + C (z - 0) = 0                               ...(3)
where  A+ 2 B - C = 0                                                                             ...(4)
                                                                   open square brackets because space plane space is space perp. space to space the space vector space straight i with hat on top space plus space 2 space straight j with hat on top space minus space straight k with hat on top close square brackets
Now plane is also normal to the vector negative straight i with hat on top space plus space straight j with hat on top space minus space 2 space straight k with hat on top as plane is perp. to line (2).
 ∴    – A + B – 2 C = 0        ... (5)
Solving (4) and (5), we get, fraction numerator straight A over denominator negative 4 plus 1 end fraction space equals fraction numerator straight B over denominator 1 plus 2 end fraction space equals fraction numerator straight C over denominator 1 plus 2 end fraction    left enclose table row 1 2 cell negative 1 end cell row cell negative 1 end cell 1 cell negative 2 end cell end table end enclose
or   fraction numerator straight A over denominator negative 3 end fraction space equals straight B over 3 space equals space straight C over 3

or  straight A over 1 space equals space fraction numerator straight B over denominator negative 1 end fraction space equals fraction numerator straight C over denominator negative 1 end fraction space equals space straight lambda space left parenthesis say right parenthesis

therefore space space space space space straight A space equals space straight lambda comma space space straight B space equals space minus straight lambda comma space space space straight C space equals space minus straight lambda
Putting these values of A, B, C in (3), we get, (x - 1) - (y – 1) – (z – 0) = 0 or x – y – z = 0
This can be written as open parentheses straight x straight i with hat on top space plus space straight y space straight j with overparenthesis on top space plus space straight z space straight k with overparenthesis on top close parentheses space open parentheses straight i with hat on top space minus space straight j with hat on top space minus space straight k with hat on top close parentheses space equals space 0
or  straight r with rightwards arrow on top. space open parentheses straight i with hat on top space minus space straight j with hat on top space minus space straight k with hat on top close parentheses space equals space 0

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180. Find the vector equation of the straight line passing through (1, 2, 3) and perpendicular to the plane straight r with rightwards arrow on top. space open parentheses straight i with hat on top space plus space 2 space straight j with hat on top space minus space 5 space straight k with hat on top close parentheses space plus space 9 space equals 0 space.
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