Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Three Dimensional Geometry

Short Answer Type

161.

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
2x + 3 y – z = 5

94 Views

162.

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
5y + 8 = 0

91 Views

Long Answer Type

163.

In the following cases, find the coordinates of the foot of the perpendicular
drawn from the origin.
2x + 3y + 4z – 12 = 0

95 Views

164. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
3 y + 4 z – 6 = 0

78 Views

Short Answer Type

165. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
x + y + z = 1

77 Views

166. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
5y + 8 = 0

97 Views

167. Find the direction cosines of the perpendicular from the origin to the plane

straight r with rightwards arrow on top. space open parentheses 6 space straight i with hat on top space minus space 3 space straight j with hat on top space minus space 2 space straight k with hat on top close parentheses space plus space 1 space equals 0

89 Views

168. Find the vector equation of the plane which is at a distance of 5 units from the origin and which is normal to the vector

2 space straight i with hat on top space plus space 6 space straight j with hat on top space minus space 3 space straight k with hat on top

81 Views

169. Find the vector equation of a plane which is at a distance of 7 units from the origin and which is normal to the vector

3 straight i with hat on top space plus space 5 space straight j with hat on top space minus space 6 space straight k with hat on top.

86 Views

Long Answer Type

170. Find the vector equation of the plane which is at a distance of $fraction numerator 6 over denominator square root of 29 end fraction$ from the origin and its normal vector from the origin is $2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space space 4 space straight k with hat on top.$ Also find its cartesian form.

Here $straight p space equals space fraction numerator 6 over denominator square root of 29 end fraction$
and $straight n with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top$

$therefore space space space space open vertical bar straight n with rightwards arrow on top close vertical bar space equals space square root of left parenthesis 2 right parenthesis squared plus left parenthesis negative 3 right parenthesis squared plus left parenthesis 4 right parenthesis squared end root space equals space square root of 4 plus 9 plus 16 end root space equals space square root of 29 therefore space space space space space space straight n with hat on top space equals space fraction numerator straight n with rightwards arrow on top over denominator open vertical bar straight n with rightwards arrow on top close vertical bar end fraction space equals space fraction numerator 1 over denominator square root of 29 end fraction left parenthesis 2 straight i with hat on top space minus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top right parenthesis$
The required equation of plane is
$straight r with rightwards arrow on top. space straight n with hat on top space equals space straight p$
or $straight r with rightwards arrow on top. space fraction numerator 1 over denominator square root of 29 end fraction space left parenthesis 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top right parenthesis space equals space fraction numerator 6 over denominator square root of 29 end fraction$
or $stack straight r. with rightwards arrow on top left parenthesis 2 space straight i with hat on top space minus space space 3 space straight j with hat on top space plus space 4 space straight k with hat on top right parenthesis space equals space 6$
Taking $straight r with rightwards arrow on top space equals space straight x space straight i with hat on top space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top comma space we space get$
$left parenthesis straight x space straight i with hat on top space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top right parenthesis. space space space open parentheses 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top close parentheses space equals space 6$
or 2x – 3 y + 4 z = 6, which is cartesian equation of plane.