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Three Dimensional Geometry

Short Answer Type

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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Long Answer Type

173. A plane meets the coordinate axes in A, B, C and (α, β, γ) is the centroid of the triangle ABC. Then, show that the equation of the plane is

straight x over straight alpha plus straight y over straight beta plus straight z over straight gamma space equals space 3.

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Short Answer Type

174. A plane meets the co-ordinate axes at A, B, C such that the centroid of the triangle ABC is the point (1, – 2, 3). Show that the equation of the plane is 6x-3y + 2z= 18.

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Long Answer Type

175. A plane meets the co-ordinate axes at A, B, C such that the centroid of triangle ABC is the point (a, b, c). Show that the equation of the plane is $straight x over straight a plus straight y over straight b plus straight z over straight c equals 3.$

Let the equation of plane be

straight x over straight alpha plus straight y over straight beta plus straight z over straight gamma equals 1

....(1)
It meets x-axis in A where y = 0, z = 0
Putting y = 0, z = 0 in (1), we get x / α = 1, ∴ x = α
∴ A is (α, 0, 0)
Similarly B, C are (0,β, 0), (0, 0, γ) respectively
∴ (a, b, c) is centroid of triangle ABC

therefore space space space straight a space equals space fraction numerator straight alpha plus 0 plus 0 over denominator 3 end fraction comma space space straight b space equals space fraction numerator 0 plus straight beta plus 0 over denominator 3 end fraction comma space space straight c space equals space fraction numerator 0 plus 0 plus straight gamma over denominator 3 end fraction. therefore space space space straight a space equals space straight alpha over 3 comma space space space straight b space equals space straight beta over 3 comma space space straight c space equals space straight gamma over 3 therefore space space space space straight alpha space equals space 3 space straight a comma space space space space space straight beta space equals space 3 space straight b comma space space space straight gamma space equals space 3 space straight c

Putting values of α, β, γ in (1), we get,

fraction numerator straight x over denominator 3 space straight a end fraction plus fraction numerator straight y over denominator 3 space straight b end fraction plus fraction numerator straight z over denominator 3 space straight c end fraction equals 1

straight x over straight a plus straight y over straight b plus straight z over straight c equals 3 comma

which is required equation of plane

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Short Answer Type

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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Long Answer Type

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 $straight r with rightwards arrow on top space equals space open parentheses straight i with hat on top plus straight j with hat on top close parentheses plus straight t open parentheses negative straight i with hat on top space plus space straight j with hat on top space minus space 2 space straight k with hat on top close parentheses$

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Short Answer Type

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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