Q1. If a line makes angles α,β,γ and δ with four diagonals of a cube, then the value of sin^2α+sin^2β+sin^2γ+sin^2δ, is
Solution
We know
cos^2α+cos^2β+cos^2γ+cos^2 δ=4/3
where α, β, γ and δ are the angles with diagonals of cube.
∴1-sin^2 α+1-sin^2 β+1-sin^2 γ+1-sin^2 δ=4/3
⟹ sin^2 α+sin^2 β+sin^2 γ+sin^2 δ=8/3
Q2.If P(x,y,z) is a point on the line segment joining Q(2,24) and R(3,5,6) such that the projections of OP on the axes are 13/5,19/5 and 26/5 respectively, then P divides QR in the ratio
Solution
The coordinate of P are
((3λ+2)/(λ+1),(5λ+2)/(λ+1),(6λ+4)/(λ+1))
Since, the projection of OP on x-axis is
(3λ+2)/(λ+1)=13/5
⟹15λ+10=13λ+13
⟹ λ=3/2
Q3. The symmetric equation of lines 3x+2y+z-5=0 and x+y-2z-3=0, is
Solution
Let a,b,c be the direction ratios of required line.
∴3a+2b+c=0 and a+b-2c=0
⟹a/(-4-1)=b/(1+6)=c/(3-2)
⟹a/(-5)=b/7=c/1
In order to find a point on the required line we put z=0 in the two given equations to obtain, 3x+2y=5 and x+y=3
∴ Coordinate of point on required line are (-1,4,0)
Hence, required line is
(x+1)/(-5)=(y-4)/7=(z-0)/1
Q4. The equation of the plane containing the two lines (x-1)/2=(y+1)/(-1)=z/3 and x/(-1)=(y-2)/3=(z+1)/(-1) is
Solution
Q5.A point moves such that the sum of its distance from points (4, 0, 0) and (-4,0,0) is 10, then the locus of the point is
Solution
Q6. If for a plane, the intercepts on the coordinate axes are 8, 4, 4 then the length of the perpendicular from the origin on to the plane is
Solution
Equation of plane isx/8+y/4+z/4=1
⟹x+2y+2z=8
Length of perpendicular from origin to x+2y+2z-8=0
=|(-8)/√(1+4+4)|=8/3
Q7.The centre and radius of the sphere x^2+y^2+z^2+3x-4z+1=0 are
Solution
The given equation of sphere is
x^2+y^2+z^2+3x-4z+1=0
∴ Coordinates of centre of sphere=(-3/2,0,2)
and radius of sphere =√(u^2+v^2+w^2-d)
=√(((9/4)+4-1)=√21/2
Q8.If P be the point (2, 6, 3), then the equation of the plane through P at right angle to OP,O being the origin, is
Solution
Distance of point P(2,6,3) from origin
OP=√((0-2)^2+(0-6)^2+(0-3)^2 )
=√(4+36+9)=7
Now, DR’s of OP=2-0,6-0,3-0=2,6,3
∴ DC’s of OP are 2/7,6/7,3/7
∴ Equation of plane in normal form is
lx+my+nz=p
⇒2/7 x+6/7 y+3/7 z=7
⇒2x+6y+3z=49
Q9.A variable plane which remains at a constant distance p from the origin cuts the coordinate axes in A,B,C. The locus of the centroid of the tetrahedron OABC is y^2 z^2+z^2 x^2+x^2 y^2=k x^2 y^2 z^2, where k is equal to
Solution
Q10. If P=(0,1,2),Q=(4,-2,1),O=(0,0,0), then ∠ POQ is equal to
Solution
Q1. If a line makes angles α,β,γ and δ with four diagonals of a cube, then the value of sin^2α+sin^2β+sin^2γ+sin^2δ, is
Solution
We know cos^2α+cos^2β+cos^2γ+cos^2 δ=4/3 where α, β, γ and δ are the angles with diagonals of cube. ∴1-sin^2 α+1-sin^2 β+1-sin^2 γ+1-sin^2 δ=4/3 ⟹ sin^2 α+sin^2 β+sin^2 γ+sin^2 δ=8/3
We know cos^2α+cos^2β+cos^2γ+cos^2 δ=4/3 where α, β, γ and δ are the angles with diagonals of cube. ∴1-sin^2 α+1-sin^2 β+1-sin^2 γ+1-sin^2 δ=4/3 ⟹ sin^2 α+sin^2 β+sin^2 γ+sin^2 δ=8/3
Q2.If P(x,y,z) is a point on the line segment joining Q(2,24) and R(3,5,6) such that the projections of OP on the axes are 13/5,19/5 and 26/5 respectively, then P divides QR in the ratio
Solution
The coordinate of P are ((3λ+2)/(λ+1),(5λ+2)/(λ+1),(6λ+4)/(λ+1)) Since, the projection of OP on x-axis is (3λ+2)/(λ+1)=13/5 ⟹15λ+10=13λ+13 ⟹ λ=3/2
The coordinate of P are ((3λ+2)/(λ+1),(5λ+2)/(λ+1),(6λ+4)/(λ+1)) Since, the projection of OP on x-axis is (3λ+2)/(λ+1)=13/5 ⟹15λ+10=13λ+13 ⟹ λ=3/2
Q3. The symmetric equation of lines 3x+2y+z-5=0 and x+y-2z-3=0, is
Solution
Let a,b,c be the direction ratios of required line. ∴3a+2b+c=0 and a+b-2c=0 ⟹a/(-4-1)=b/(1+6)=c/(3-2) ⟹a/(-5)=b/7=c/1 In order to find a point on the required line we put z=0 in the two given equations to obtain, 3x+2y=5 and x+y=3 ∴ Coordinate of point on required line are (-1,4,0) Hence, required line is (x+1)/(-5)=(y-4)/7=(z-0)/1
Let a,b,c be the direction ratios of required line. ∴3a+2b+c=0 and a+b-2c=0 ⟹a/(-4-1)=b/(1+6)=c/(3-2) ⟹a/(-5)=b/7=c/1 In order to find a point on the required line we put z=0 in the two given equations to obtain, 3x+2y=5 and x+y=3 ∴ Coordinate of point on required line are (-1,4,0) Hence, required line is (x+1)/(-5)=(y-4)/7=(z-0)/1
Q4. The equation of the plane containing the two lines (x-1)/2=(y+1)/(-1)=z/3 and x/(-1)=(y-2)/3=(z+1)/(-1) is
Solution
Q5.A point moves such that the sum of its distance from points (4, 0, 0) and (-4,0,0) is 10, then the locus of the point is
Solution
Q6. If for a plane, the intercepts on the coordinate axes are 8, 4, 4 then the length of the perpendicular from the origin on to the plane is
Solution
Equation of plane isx/8+y/4+z/4=1 ⟹x+2y+2z=8 Length of perpendicular from origin to x+2y+2z-8=0 =|(-8)/√(1+4+4)|=8/3
Equation of plane isx/8+y/4+z/4=1 ⟹x+2y+2z=8 Length of perpendicular from origin to x+2y+2z-8=0 =|(-8)/√(1+4+4)|=8/3
Q7.The centre and radius of the sphere x^2+y^2+z^2+3x-4z+1=0 are
Solution
The given equation of sphere is x^2+y^2+z^2+3x-4z+1=0 ∴ Coordinates of centre of sphere=(-3/2,0,2) and radius of sphere =√(u^2+v^2+w^2-d) =√(((9/4)+4-1)=√21/2
The given equation of sphere is x^2+y^2+z^2+3x-4z+1=0 ∴ Coordinates of centre of sphere=(-3/2,0,2) and radius of sphere =√(u^2+v^2+w^2-d) =√(((9/4)+4-1)=√21/2
Q8.If P be the point (2, 6, 3), then the equation of the plane through P at right angle to OP,O being the origin, is
Solution
Distance of point P(2,6,3) from origin OP=√((0-2)^2+(0-6)^2+(0-3)^2 ) =√(4+36+9)=7 Now, DR’s of OP=2-0,6-0,3-0=2,6,3 ∴ DC’s of OP are 2/7,6/7,3/7 ∴ Equation of plane in normal form is lx+my+nz=p ⇒2/7 x+6/7 y+3/7 z=7 ⇒2x+6y+3z=49
Distance of point P(2,6,3) from origin OP=√((0-2)^2+(0-6)^2+(0-3)^2 ) =√(4+36+9)=7 Now, DR’s of OP=2-0,6-0,3-0=2,6,3 ∴ DC’s of OP are 2/7,6/7,3/7 ∴ Equation of plane in normal form is lx+my+nz=p ⇒2/7 x+6/7 y+3/7 z=7 ⇒2x+6y+3z=49
Q9.A variable plane which remains at a constant distance p from the origin cuts the coordinate axes in A,B,C. The locus of the centroid of the tetrahedron OABC is y^2 z^2+z^2 x^2+x^2 y^2=k x^2 y^2 z^2, where k is equal to
Solution
Q10. If P=(0,1,2),Q=(4,-2,1),O=(0,0,0), then ∠ POQ is equal to
Solution
