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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.

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Short Answer Type

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

The equation of plane is

straight r with rightwards arrow on top space equals space straight i with hat on top space 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

straight x straight i with hat on top space space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top space equals space open parentheses 1 plus straight lambda space plus space straight mu close parentheses space straight i with hat on top space plus space open parentheses negative 1 plus straight lambda space minus space 2 space straight mu close parentheses straight j with hat on top space plus space open parentheses straight lambda plus space 3 space straight mu close parentheses space straight k with hat on top

Equating the coefficients of

straight i with hat on top comma space straight j with hat on top comma space straight k with hat on top comma

we get,
x = 1 + λ + μ    ....(1)
y = – 1 + λ – 2 μ    ...(2)
z = λ + 3 μ    ...(3)
We are to eliminate λ and μ from (1), (2), (3).
Subtracting (2) from (1), we get,
x – y = 2 + 3 μ    ... (4)Subtracting (3) from (1), we get,
x – z = 1 – 2 μ    ... (5)
Multiplying (4) by 2 and (5) by 3 , we get,
2 x – 2 y = 4 + 6 μ    ...(6)
3 x – 3 z = 3 – 6 μ    ...(7)
Adding (6) and (7), we get, 5 x – 2 y – 3 z = 7
This can be written as

open parentheses straight x space straight i with hat on top space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top close parentheses. space open parentheses 5 space straight i with hat on top space space minus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top close parentheses space equals space 7

straight r with rightwards arrow on top. space open parentheses 5 space straight i with hat on top space minus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top close parentheses space equals space 7

which is the required equation.

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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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