Q1. The foot of the perpendicular from (2,4,-1) to the line
Solution
In a given options, only option (a) satisfies the given equation of line.
Q2.An equation of the line passing through 3i ̂-5j ̂+7k ̂ and perpendicular to the plane 3x-4y=5z=8 is
Solution
Equation of the give plane can be writer as
(3i ̂-4j ̂+5k ̂ )∙(x i ̂+y j ̂+zk ̂ )=8
So, that the normal to the given plane is 3i ̂-4j ̂+5k ̂ and the required line being perpendicular to the plane is parallel to this normal and since, it passes through 3i ̂-5j ̂+7k ̂, its equation is
r ⃗=3i ̂-5j ̂+7k ̂+λ(3i ̂-4j ̂+5k ̂)
Where λ is a parameter
Since, this lie passes through the vector 3i ̂-5j ̂+7k ̂ ie, the point (3,-5,7) and is parallel to 3i ̂-4j ̂+5k ̂, its direction ratios are 3,-4,5
Its cartesian equation is (x-3)/3=(y+5)/(-4)=(z-7)/5
Q3. The position vector of the point in which the line joining the points i ̂-2j ̂+k ̂ and 3k ̂-2j ̂ cuts the plane through the origin and the points 4j ̂ and 2j ̂+k ̂, is
Solution
The vector equation of the line joining the points i ̂-2j ̂+k ̂ and -2j ̂+3k ̂ is
r ⃗=(i ̂-2j ̂+k ̂ )+λ(-i ̂+2k ̂ ) …(i)
Using r ⃗∙(a ⃗×b ⃗+b ⃗×c ⃗+c ⃗×a ⃗ )=[a ⃗ b ⃗ c ⃗ ] the vector equation of the plane through the origin, 4j ̂ and 2i ̂+k ̂
r ⃗∙(4i ̂-8k ̂ )=0 …(ii)
The position vector of any point on (i) is
(i ̂-2j ̂+k ̂ )+λ(-i ̂+2k ̂ )
If it lies on (ii), then
{(i ̂-2j ̂+k ̂ )+λ(-i ̂+2k ̂ )}∙(4i ̂-8k ̂ )=0
⇒-4-20λ=0⇒λ=-1/5
Putting the value of λ in (i ̂-2j ̂+k ̂ )+λ(-i ̂+2k ̂ ), we get the position vector of the required point as 1/5 (6i ̂-10j ̂+3k ̂ )
Q4. If the position vectors of the points A and B are 3i ̂+j ̂+2k ̂ and i ̂-2j ̂-4k ̂ respectively, then the equation of the plane through B and perpendicular to AB is
Solution
We have, A ⃗B=-2i ̂-3j ̂-6k ̂
So, vector equation of the plane is
{r ⃗-(i ̂-2j ̂-4k ̂ )}∙A ⃗B=0
⇒r ⃗∙(-2i ̂-3j ̂-6k ̂ )=(i ̂-2j ̂-4k ̂ )∙(-2i ̂-3j ̂-6k ̂ )
⇒-2x-3y-6z=-2+6+24⇒2x+3y+6x+28=0
Q5.If P is a point in space such that OP=12 and O ⃗P is inclined at angles of 45° and 60° with OX and OY respectively, then the position vector of P is
Solution
Let l,m,n be the direction cosines of O ⃗P. It is given that l=45°=1/√2 and m=cos〖60°〗=1/2
∴l^2+m^2+n^2=1⇒1/2+1/4+n^2=1⇒n=±1/2
Now, r ⃗=|r ⃗ |(l i ̂+m j ̂+n k ̂ )
⇒r ⃗=12(1/√2 i ̂+1/2 j ̂±1/2 k ̂ )=6√2 i ̂+6j ̂±6k ̂
Q6. If the straight lines x=1+s,y=-3-λs,z=1+λs and
x=t/2,y=1+t,z=2-t with parameters s and t respectively, are coplanar, then λ
Equals
Solution
Q7.A line makes an obtuse angle with the positive x-axis and angles π/4 and π/3 with the positive y and z-axes respectively. Its direction cosine are
Solution
Let,m=cos〖Ï€/4=1/√2〗
and n=cos〖Ï€/3=1/2〗
∵ l^2+m^2+n^2=1
⟹l=√(1-(m^2+n^2 ) )
=√(1-(1/2+1/4) )
=√(1-3/4)
⟹l=±1/2
Since, line makes an obtuse angle, so we take
l=-1/2
∴Direction cosines are-1/2,1/√2,1/2
Q8.The equation |x|=p,|y|=p,|z|=p in xyz space represent
Solution
Since, we are given the equal intercept of the coordinate axes ie,|x|=|y|=|z|=p
Therefore, it make a cube
Q9.The equation of the line of intersection of the planes x+2y+z=3 and 6x+8y+3z=13 can be written as
Solution
A
Q10. The line (x-1)/2=(y-2)/3=(z-3)/4 meets the plane 2x+3y-z=-4 in the point
Solution
Any point on the line
Q1. The foot of the perpendicular from (2,4,-1) to the line
Solution
In a given options, only option (a) satisfies the given equation of line.
In a given options, only option (a) satisfies the given equation of line.
Q2.An equation of the line passing through 3i ̂-5j ̂+7k ̂ and perpendicular to the plane 3x-4y=5z=8 is
Solution
Equation of the give plane can be writer as (3i ̂-4j ̂+5k ̂ )∙(x i ̂+y j ̂+zk ̂ )=8 So, that the normal to the given plane is 3i ̂-4j ̂+5k ̂ and the required line being perpendicular to the plane is parallel to this normal and since, it passes through 3i ̂-5j ̂+7k ̂, its equation is r ⃗=3i ̂-5j ̂+7k ̂+λ(3i ̂-4j ̂+5k ̂) Where λ is a parameter Since, this lie passes through the vector 3i ̂-5j ̂+7k ̂ ie, the point (3,-5,7) and is parallel to 3i ̂-4j ̂+5k ̂, its direction ratios are 3,-4,5 Its cartesian equation is (x-3)/3=(y+5)/(-4)=(z-7)/5
Equation of the give plane can be writer as (3i ̂-4j ̂+5k ̂ )∙(x i ̂+y j ̂+zk ̂ )=8 So, that the normal to the given plane is 3i ̂-4j ̂+5k ̂ and the required line being perpendicular to the plane is parallel to this normal and since, it passes through 3i ̂-5j ̂+7k ̂, its equation is r ⃗=3i ̂-5j ̂+7k ̂+λ(3i ̂-4j ̂+5k ̂) Where λ is a parameter Since, this lie passes through the vector 3i ̂-5j ̂+7k ̂ ie, the point (3,-5,7) and is parallel to 3i ̂-4j ̂+5k ̂, its direction ratios are 3,-4,5 Its cartesian equation is (x-3)/3=(y+5)/(-4)=(z-7)/5
Q3. The position vector of the point in which the line joining the points i ̂-2j ̂+k ̂ and 3k ̂-2j ̂ cuts the plane through the origin and the points 4j ̂ and 2j ̂+k ̂, is
Solution
The vector equation of the line joining the points i ̂-2j ̂+k ̂ and -2j ̂+3k ̂ is r ⃗=(i ̂-2j ̂+k ̂ )+λ(-i ̂+2k ̂ ) …(i) Using r ⃗∙(a ⃗×b ⃗+b ⃗×c ⃗+c ⃗×a ⃗ )=[a ⃗ b ⃗ c ⃗ ] the vector equation of the plane through the origin, 4j ̂ and 2i ̂+k ̂ r ⃗∙(4i ̂-8k ̂ )=0 …(ii) The position vector of any point on (i) is (i ̂-2j ̂+k ̂ )+λ(-i ̂+2k ̂ ) If it lies on (ii), then {(i ̂-2j ̂+k ̂ )+λ(-i ̂+2k ̂ )}∙(4i ̂-8k ̂ )=0 ⇒-4-20λ=0⇒λ=-1/5 Putting the value of λ in (i ̂-2j ̂+k ̂ )+λ(-i ̂+2k ̂ ), we get the position vector of the required point as 1/5 (6i ̂-10j ̂+3k ̂ )
The vector equation of the line joining the points i ̂-2j ̂+k ̂ and -2j ̂+3k ̂ is r ⃗=(i ̂-2j ̂+k ̂ )+λ(-i ̂+2k ̂ ) …(i) Using r ⃗∙(a ⃗×b ⃗+b ⃗×c ⃗+c ⃗×a ⃗ )=[a ⃗ b ⃗ c ⃗ ] the vector equation of the plane through the origin, 4j ̂ and 2i ̂+k ̂ r ⃗∙(4i ̂-8k ̂ )=0 …(ii) The position vector of any point on (i) is (i ̂-2j ̂+k ̂ )+λ(-i ̂+2k ̂ ) If it lies on (ii), then {(i ̂-2j ̂+k ̂ )+λ(-i ̂+2k ̂ )}∙(4i ̂-8k ̂ )=0 ⇒-4-20λ=0⇒λ=-1/5 Putting the value of λ in (i ̂-2j ̂+k ̂ )+λ(-i ̂+2k ̂ ), we get the position vector of the required point as 1/5 (6i ̂-10j ̂+3k ̂ )
Q4. If the position vectors of the points A and B are 3i ̂+j ̂+2k ̂ and i ̂-2j ̂-4k ̂ respectively, then the equation of the plane through B and perpendicular to AB is
Solution
We have, A ⃗B=-2i ̂-3j ̂-6k ̂ So, vector equation of the plane is {r ⃗-(i ̂-2j ̂-4k ̂ )}∙A ⃗B=0 ⇒r ⃗∙(-2i ̂-3j ̂-6k ̂ )=(i ̂-2j ̂-4k ̂ )∙(-2i ̂-3j ̂-6k ̂ ) ⇒-2x-3y-6z=-2+6+24⇒2x+3y+6x+28=0
We have, A ⃗B=-2i ̂-3j ̂-6k ̂ So, vector equation of the plane is {r ⃗-(i ̂-2j ̂-4k ̂ )}∙A ⃗B=0 ⇒r ⃗∙(-2i ̂-3j ̂-6k ̂ )=(i ̂-2j ̂-4k ̂ )∙(-2i ̂-3j ̂-6k ̂ ) ⇒-2x-3y-6z=-2+6+24⇒2x+3y+6x+28=0
Q5.If P is a point in space such that OP=12 and O ⃗P is inclined at angles of 45° and 60° with OX and OY respectively, then the position vector of P is
Solution
Let l,m,n be the direction cosines of O ⃗P. It is given that l=45°=1/√2 and m=cos〖60°〗=1/2 ∴l^2+m^2+n^2=1⇒1/2+1/4+n^2=1⇒n=±1/2 Now, r ⃗=|r ⃗ |(l i ̂+m j ̂+n k ̂ ) ⇒r ⃗=12(1/√2 i ̂+1/2 j ̂±1/2 k ̂ )=6√2 i ̂+6j ̂±6k ̂
Let l,m,n be the direction cosines of O ⃗P. It is given that l=45°=1/√2 and m=cos〖60°〗=1/2 ∴l^2+m^2+n^2=1⇒1/2+1/4+n^2=1⇒n=±1/2 Now, r ⃗=|r ⃗ |(l i ̂+m j ̂+n k ̂ ) ⇒r ⃗=12(1/√2 i ̂+1/2 j ̂±1/2 k ̂ )=6√2 i ̂+6j ̂±6k ̂
Q6. If the straight lines x=1+s,y=-3-λs,z=1+λs and
x=t/2,y=1+t,z=2-t with parameters s and t respectively, are coplanar, then λ
Equals
Solution
Q7.A line makes an obtuse angle with the positive x-axis and angles π/4 and π/3 with the positive y and z-axes respectively. Its direction cosine are
Solution
Let,m=cos〖Ï€/4=1/√2〗 and n=cos〖Ï€/3=1/2〗 ∵ l^2+m^2+n^2=1 ⟹l=√(1-(m^2+n^2 ) ) =√(1-(1/2+1/4) ) =√(1-3/4) ⟹l=±1/2 Since, line makes an obtuse angle, so we take l=-1/2 ∴Direction cosines are-1/2,1/√2,1/2
Let,m=cos〖Ï€/4=1/√2〗 and n=cos〖Ï€/3=1/2〗 ∵ l^2+m^2+n^2=1 ⟹l=√(1-(m^2+n^2 ) ) =√(1-(1/2+1/4) ) =√(1-3/4) ⟹l=±1/2 Since, line makes an obtuse angle, so we take l=-1/2 ∴Direction cosines are-1/2,1/√2,1/2
Q8.The equation |x|=p,|y|=p,|z|=p in xyz space represent
Solution
Since, we are given the equal intercept of the coordinate axes ie,|x|=|y|=|z|=p Therefore, it make a cube
Since, we are given the equal intercept of the coordinate axes ie,|x|=|y|=|z|=p Therefore, it make a cube
Q9.The equation of the line of intersection of the planes x+2y+z=3 and 6x+8y+3z=13 can be written as
Solution
A
A
Q10. The line (x-1)/2=(y-2)/3=(z-3)/4 meets the plane 2x+3y-z=-4 in the point
Solution
Any point on the line
Any point on the line
