Mathematics Question Papers for IIT JEE Exam on “Three Dimensional Geometry – Equation of a Line in Space”.
1. Find the vector equation of the line which is passing through the point (2,-3,5) and parallel to the vector (3hat{i}+4hat{j}-2hat{k}).
a) ((2+3λ) hat{i}+(4λ+3) hat{j}+(5-λ)hat{k})
b) ((9+3λ) hat{i}+(λ-3) hat{j}+(5-2λ)hat{k})
c) ((2+3λ) hat{i}+(4λ-3) hat{j}+(5-2λ)hat{k})
d) ((7+λ) hat{i}+(4λ+3) hat{j}+(5-2λ)hat{k})
Answer: c
Clarification: Given that the line is passing through the point (2,-3,5). Therefore, the position vector of the line is (vec{a}=2hat{i}-3hat{j}+5hat{k}).
Also given that, the line is parallel to a vector (vec{b}=3hat{i}+4hat{j}-2hat{k}).
We know that, the equation of line passing through a point and parallel to vector is given by (vec{r}=vec{a}+λvec{b}), where λ is a constant.
∴(vec{r}=2hat{i}-3hat{j}+5hat{k}+λ(3hat{i}+4hat{j}-2hat{k}))
=((2+3λ) hat{i}+(4λ-3) hat{j}+(5-2λ)hat{k})
2. If the line is passing through the points ((x_1, y_1, z_1)) and has direction cosines l, m, n of the line, then which of the following is the cartesian equation of the line?
a) (frac{x-x_1}{l}=frac{y-y_1}{m}=frac{z-z_1}{n})
b) (frac{x-x_1}{n}=frac{y-y_1}{m}=frac{z-z_1}{l})
c) (frac{x+x_1}{n}=frac{y+y_1}{m}=frac{z-z_1}{l})
d) (frac{x+x_1}{l}=frac{y+y_1}{m}=frac{z+z_1}{n})
Answer: a
Clarification: If the line is passing through the points (x1, y1, z1) and has direction cosines l,m,n of the line, then the cartesian equation of the line is given by
(frac{x-x_1}{l}=frac{y-y_1}{m}=frac{z-z_1}{n}).
3. If a line is passing through two points (A(x_1,y_1,z_1)) and (B(x_2,y_2,z_2)) then which of the following is the vector equation of the line?
a) (vec{r}=vec{a}+λ(vec{b}+vec{a}))
b) (vec{r}=vec{a}+λ(vec{a}-vec{b}))
c) (vec{r}=λvec{a}+(vec{b}-vec{a}))
d) (vec{r}=vec{a}+λ(vec{b}-vec{a}))
Answer: d
Clarification: Let (vec{a} ,and, vec{b}) are the position vectors of two points (A(x_1,y_1,z_1)) and (B(x_2,y_2,z_2))
Then the vector equation of the line is given by the formula passing through two points will be given by
(vec{r}=vec{a}+λ(vec{b}-vec{a})), λ∈R
4. Find the vector equation of a line passing through two points (-5,3,1) and (4,-3,2).
a) ((-5+λ) hat{i}+(3+λ)hat{j}+(1-λ) hat{k})
b) ((-5+λ) hat{i}+(3+6λ)hat{j}+(1+λ) hat{k})
c) ((5+7λ) hat{i}+(8+6λ)hat{j}+(3-5λ) hat{k})
d) ((-5+9λ) hat{i}+(3-6λ)hat{j}+(1+λ) hat{k})
Answer: d
Clarification: Consider the points A(-5,3,1) and B(4,-3,2)
Let (vec{a} ,and, vec{b}) be the position vectors of the points A and B.
∴(vec{a}=-5hat{i}+3hat{j}+hat{k})
(vec{b}=4hat{i}-3hat{j}+2hat{k})
∴(vec{r}=-5hat{i}+3hat{j}+hat{k}+λ(4hat{i}-3hat{j}+2hat{k}-(-5hat{i}+3hat{j}+hat{k})))
=-(5hat{i}+3hat{j}+hat{k}+λ(9hat{i}-6hat{j}+hat{k}))
=((-5+9λ) hat{i}+(3-6λ)hat{j}+(1+λ) hat{k})
5. Find the cartesian equation of a line passing through two points (1,-9,8) and (4,-1,6).
a) (frac{x+1}{3}=frac{y-9}{8}=frac{-z-8}{2})
b) (frac{x-1}{3}=frac{y+9}{8}=frac{z-8}{-2})
c) (frac{x-1}{7}=frac{y+9}{-2}=frac{z-8}{5})
d) (frac{2x-1}{3}=frac{6y+9}{8}=frac{4z-8}{-2})
Answer: b
Clarification: The position vector for the point A(1,-9,8) and B(4,-1,6)
(vec{a}=hat{i}-9hat{j}+8hat{k})
(vec{b}=4hat{i}-hat{j}+6hat{k})
∴(vec{r}=vec{a}+λ(vec{b}-vec{a}))
The above vector equation can be expressed in cartesian form as:
(frac{x-x_1}{x_2-x_1}=frac{y-y_1}{y_2-y_1}=frac{z-z_1}{z_2-z_1})
∴ The cartesian equation for the given line is (frac{x-1}{3}=frac{y+9}{8}=frac{z-8}{-2})
6. Find the cartesian equation of the line which is passing through the point (5,-6,1) and parallel to the vector (5hat{i}+2hat{j}-3hat{k}).
a) (frac{x-5}{5}=frac{y-4}{6}=frac{z+1}{-3})
b) (frac{x-5}{5}=frac{z+6}{2}=frac{y-1}{3})
c) (frac{x+5}{4}=frac{y-8}{2}=frac{z-1}{-3})
d) (frac{x-5}{5}=frac{y+6}{2}=frac{z-1}{-3})
Answer: d
Clarification: The equation of a line passing through a point and parallel to a vector is given by
(vec{r}=vec{a}+λvec{b})
(vec{a}) is the position vector of the given point ∴(vec{a}=5hat{i}-6hat{j}+hat{k})
(vec{b}=5hat{i}+2hat{j}-3hat{k}).
(vec{r}=5hat{i}-6hat{j}+hat{k}+λ(5hat{i}+2hat{j}-3hat{k}))
(xhat{i}+yhat{j}+zhat{k}=(5+5λ) hat{i}+(2λ-6) hat{j}+(1-3λ) hat{k})
∴(frac{x-5}{5}=frac{y+6}{2}=frac{z-1}{-3})
is the cartesian equation of the given line.
7. Find the cartesian equation of a line passing through two points (8,-5,7) and (7,1,4).
a) (frac{x-8}{1}=frac{y+5}{-6}=frac{z-7}{3})
b) (frac{2x-8}{-1}=frac{3y+5}{6}=frac{4z-7}{-3})
c) (frac{x-8}{-1}=frac{y+5}{6}=frac{z-7}{-3})
d) (frac{x-8}{-2}=frac{y+5}{5}=frac{z-7}{-7})
Answer: c
Clarification: Consider A(8,-5,7) and B(7,1,4)
i.e. ((x_1,y_1,z_1))=(8,-5,7) and ((x_2,y_2,z_3))=(7,1,4)
The cartesian equation for a line passing through two points is given by
(frac{x-x_1}{x_2-x_1}=frac{y-y_1}{y_2-y_1}=frac{z-z_1}{z_2-z_1})
∴ the cartesian equation for the given line is (frac{x-8}{-1}=frac{y+5}{6}=frac{z-7}{-3})
8. Find the vector equation of a line passing through two points (1,0,4) and (6,-3,1).
a) ((1+5λ) hat{i}-λhat{j}+(4-3λ) hat{k})
b) ((1+5λ) hat{i}-3λhat{j}+(7-3λ) hat{k})
c) ((1+λ) hat{i}+λhat{j}+(8-3λ) hat{k})
d) ((1+5λ) hat{i}-3λhat{j}+(4-3λ) hat{k})
Answer: d
Clarification: Consider the points A(1,0,4) and B(6,-3,1)
Let (vec{a} ,and ,vec{b}) be the position vectors of the points A and B.
∴(vec{a}=hat{i}+4hat{k})
(vec{b}=6hat{i}-3hat{j}+hat{k})
∴(vec{r}=hat{i}+4hat{k}+λ(6hat{i}-3hat{j}+hat{k}-(hat{i}+4hat{k})))
=(hat{i}+4hat{k}+λ(5hat{i}-3hat{j}-3hat{k}))
=((1+5λ) hat{i}-3λhat{j}+(4-3λ) hat{k})
9. Find the vector equation of the line which is passing through the point (1, -4, 4) and parallel to the vector (2hat{i}-5hat{j}+2hat{k}).
a) ((1+2λ) hat{i}-(4+5λ) hat{j}+(4+2λ) hat{j})
b) ((1+2λ) hat{i}-(4+5λ) hat{j}+2λ hat{j})
c) ((1-2λ) hat{i}+(4+5λ) hat{j}+(4+2λ) hat{j})
d) ((8+λ) hat{i}-(4-5λ) hat{j}+(7-λ) hat{j})
Answer: a
Clarification: We know that, the equation of a vector passing through a point and parallel to another vector is given by (vec{r}=vec{a}+λvec{b}), where λ is a constant.
The position of vector of the point (1,-4,4) is given by (vec{a}=hat{i}-4hat{j}+4hat{k})
And (vec{b}=2hat{i}-5hat{j}+2hat{k})
∴(vec{r}=vec{a}+λvec{b})
=(hat{i}-4hat{j}+4hat{k}+λ(2hat{i}-5hat{j}+2hat{k}))
=((1+2λ) hat{i}-(4+5λ) hat{j}+(4+2λ) hat{j})
10. Find the cartesian equation of the line which is passing through the point (-1, -8, 5) and parallel to the vector (7hat{i}-3hat{j}-3hat{k}).
a) (frac{x+1}{-1}=frac{y+8}{-8}=frac{z-5}{5})
b) (frac{x+1}{7}=frac{y+8}{-3}=frac{z-5}{-3})
c) (frac{x-7}{7}=frac{y+8}{-3}=frac{z-3}{-3})
d) (frac{x+1}{7}=frac{y+8}{-8}=frac{z+5}{3})
Answer: b
Clarification: The position vector of the given point is (vec{a}=-hat{i}-8hat{j}+5hat{k})
The vector which is parallel to the given line is (7hat{i}-3hat{j}-3hat{k})
We know that, (vec{r}=vec{a}+λvec{b})
∴(xhat{i}+yhat{j}+zhat{k}=-hat{i}-8hat{j}+5hat{k}+λ(7hat{i}-3hat{j}-3hat{k}))
=((-1+7λ) hat{i}+(-8-3λ) hat{j}+(5-3λ) hat{j})
⇒(frac{x+1}{7}=frac{y+8}{-3}=frac{z-5}{-3}).
Mathematics Question Papers for IIT JEE Exam,