Q1. If projection of a line on x,y and z-axes are 6,2 and 3 respectively, then direction cosines of the line is
Solution
Direction cosines
=(6/√(36+4+9),2/√(36+4+9),3/√(36+4+9))
=(6/7,2/7,3/7)
Q2.The shortest distance between the lines (x-3)/3=(y-8)/(-1)=(z-3)/1 and (x+3)/(-3)=(y+7)/2=(z-6)/4 is
Q3. The equation of the plane through the line of intersection of planes ax+by+cz+d=0,a^'x+b^'y+c^'z+d^'=0 and parallel to the line y=0,z=0 is
Solution
Q4. The vector equation of the plane passing through the origin and the line of intersection of the plane r ⃗∙a ⃗=λ and r ⃗∙b ⃗=μ is
Q6. If the direction cosines of a line are (1/c,1/c,1/c), then
Solution
Since,DC^' sof a line are(1/c,1/c,1/c)
∴ (1/c)^2+(1/c)^2+(1/c)^2=1
⟹ c^2=3⟹c=± √3
Q7.The xy-plane divides the line joining the points (-1,3,4)and (2,-5,6)
Solution
Suppose xy-plane divides at the line joining the given points in the ratio λ : 1 . The
coordinate of the points of division are[(2λ-1)/(λ+1),(-5λ+3)/(λ+1),(6λ+4)/(λ+1)] Since the point lies
on the xy-plane
∴(6λ+4)/(λ+1)=0⟹λ=(-2)/3
Q8.The equation of the plane containing the line
(x-x1)/l=(y-y1)/m=(z-z1)/n is
Solution
Required plane contains the given line, so normal to the plane must be perpendicular to the line and the condition for the same is al+bm+cn=0.
Q9.If x coordinate of a point P of line joining the points Q(2,2,1) and R(5,2,-2) is 4, then the z coordinate of P is
Solution
Suppose P divides QR in the ratio λ:1. Then, coordinates of P are ((5λ+2)/(λ+1),(2λ+2)/(λ+1),(-2λ+1)/(λ+1))
Since, the x coordinates of P is 4
ie,(5λ+2)/(λ+1)=4 ⇒ λ=2
So, z coordinate of P is (-2λ+1)/(λ+1)=(-4+1)/(2+1)=-1
Q1. If projection of a line on x,y and z-axes are 6,2 and 3 respectively, then direction cosines of the line is
Solution
Direction cosines =(6/√(36+4+9),2/√(36+4+9),3/√(36+4+9)) =(6/7,2/7,3/7)
Direction cosines =(6/√(36+4+9),2/√(36+4+9),3/√(36+4+9)) =(6/7,2/7,3/7)
Q2.The shortest distance between the lines (x-3)/3=(y-8)/(-1)=(z-3)/1 and (x+3)/(-3)=(y+7)/2=(z-6)/4 is
Q3. The equation of the plane through the line of intersection of planes ax+by+cz+d=0,a^'x+b^'y+c^'z+d^'=0 and parallel to the line y=0,z=0 is
Solution
Q4. The vector equation of the plane passing through the origin and the line of intersection of the plane r ⃗∙a ⃗=λ and r ⃗∙b ⃗=μ is
Q6. If the direction cosines of a line are (1/c,1/c,1/c), then
Solution
Since,DC^' sof a line are(1/c,1/c,1/c) ∴ (1/c)^2+(1/c)^2+(1/c)^2=1 ⟹ c^2=3⟹c=± √3
Since,DC^' sof a line are(1/c,1/c,1/c) ∴ (1/c)^2+(1/c)^2+(1/c)^2=1 ⟹ c^2=3⟹c=± √3
Q7.The xy-plane divides the line joining the points (-1,3,4)and (2,-5,6)
Solution
Suppose xy-plane divides at the line joining the given points in the ratio λ : 1 . The coordinate of the points of division are[(2λ-1)/(λ+1),(-5λ+3)/(λ+1),(6λ+4)/(λ+1)] Since the point lies on the xy-plane ∴(6λ+4)/(λ+1)=0⟹λ=(-2)/3
Suppose xy-plane divides at the line joining the given points in the ratio λ : 1 . The coordinate of the points of division are[(2λ-1)/(λ+1),(-5λ+3)/(λ+1),(6λ+4)/(λ+1)] Since the point lies on the xy-plane ∴(6λ+4)/(λ+1)=0⟹λ=(-2)/3
Q8.The equation of the plane containing the line
(x-x1)/l=(y-y1)/m=(z-z1)/n is
Solution
Required plane contains the given line, so normal to the plane must be perpendicular to the line and the condition for the same is al+bm+cn=0.
Required plane contains the given line, so normal to the plane must be perpendicular to the line and the condition for the same is al+bm+cn=0.
Q9.If x coordinate of a point P of line joining the points Q(2,2,1) and R(5,2,-2) is 4, then the z coordinate of P is
Solution
Suppose P divides QR in the ratio λ:1. Then, coordinates of P are ((5λ+2)/(λ+1),(2λ+2)/(λ+1),(-2λ+1)/(λ+1)) Since, the x coordinates of P is 4 ie,(5λ+2)/(λ+1)=4 ⇒ λ=2 So, z coordinate of P is (-2λ+1)/(λ+1)=(-4+1)/(2+1)=-1
Suppose P divides QR in the ratio λ:1. Then, coordinates of P are ((5λ+2)/(λ+1),(2λ+2)/(λ+1),(-2λ+1)/(λ+1)) Since, the x coordinates of P is 4 ie,(5λ+2)/(λ+1)=4 ⇒ λ=2 So, z coordinate of P is (-2λ+1)/(λ+1)=(-4+1)/(2+1)=-1