# Mathematics Questions and Answers – Three Dimensional Geometry – Equation of a Line in Space

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This set of Mathematics Question Papers for IIT JEE Exam focuses 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 $$3\hat{i}+4\hat{j}-2\hat{k}$$.
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) $$(7+λ) \hat{i}+(4λ+3) \hat{j}+(5-2λ)\hat{k}$$

Explanation: Given that the line is passing through the point (2,-3,5). Therefore, the position vector of the line is $$\vec{a}=2\hat{i}-3\hat{j}+5\hat{k}$$.
Also given that, the line is parallel to a vector $$\vec{b}=3\hat{i}+4\hat{j}-2\hat{k}$$.
We know that, the equation of line passing through a point and parallel to vector is given by $$\vec{r}=\vec{a}+λ\vec{b}$$, where λ is a constant.
∴$$\vec{r}=2\hat{i}-3\hat{j}+5\hat{k}+λ(3\hat{i}+4\hat{j}-2\hat{k})$$
=$$(2+3λ) \hat{i}+(4λ-3) \hat{j}+(5-2λ)\hat{k}$$

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) $$\frac{x+x_1}{l}=\frac{y+y_1}{m}=\frac{z+z_1}{n}$$

Explanation: 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) $$\vec{r}=λ\vec{a}+(\vec{b}-\vec{a})$$
d) $$\vec{r}=\vec{a}+λ(\vec{b}-\vec{a})$$

Explanation: 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) $$(-5+9λ) \hat{i}+(3-6λ)\hat{j}+(1+λ) \hat{k}$$

Explanation: 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.
∴$$\vec{a}=-5\hat{i}+3\hat{j}+\hat{k}$$
$$\vec{b}=4\hat{i}-3\hat{j}+2\hat{k}$$
∴$$\vec{r}=-5\hat{i}+3\hat{j}+\hat{k}+λ(4\hat{i}-3\hat{j}+2\hat{k}-(-5\hat{i}+3\hat{j}+\hat{k}))$$
=-$$5\hat{i}+3\hat{j}+\hat{k}+λ(9\hat{i}-6\hat{j}+\hat{k})$$
=$$(-5+9λ) \hat{i}+(3-6λ)\hat{j}+(1+λ) \hat{k}$$

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) $$\frac{2x-1}{3}=\frac{6y+9}{8}=\frac{4z-8}{-2}$$

Explanation: The position vector for the point A(1,-9,8) and B(4,-1,6)
$$\vec{a}=\hat{i}-9\hat{j}+8\hat{k}$$
$$\vec{b}=4\hat{i}-\hat{j}+6\hat{k}$$
∴$$\vec{r}=\vec{a}+λ(\vec{b}-\vec{a})$$
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 $$5\hat{i}+2\hat{j}-3\hat{k}$$.
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) $$\frac{x-5}{5}=\frac{y+6}{2}=\frac{z-1}{-3}$$

Explanation: The equation of a line passing through a point and parallel to a vector is given by
$$\vec{r}=\vec{a}+λ\vec{b}$$
$$\vec{a}$$ is the position vector of the given point ∴$$\vec{a}=5\hat{i}-6\hat{j}+\hat{k}$$
$$\vec{b}=5\hat{i}+2\hat{j}-3\hat{k}$$.
$$\vec{r}=5\hat{i}-6\hat{j}+\hat{k}+λ(5\hat{i}+2\hat{j}-3\hat{k})$$
$$x\hat{i}+y\hat{j}+z\hat{k}=(5+5λ) \hat{i}+(2λ-6) \hat{j}+(1-3λ) \hat{k}$$
∴$$\frac{x-5}{5}=\frac{y+6}{2}=\frac{z-1}{-3}$$
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) $$\frac{x-8}{-2}=\frac{y+5}{5}=\frac{z-7}{-7}$$

Explanation: 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) $$(1+5λ) \hat{i}-λ\hat{j}+(4-3λ) \hat{k}$$
b) $$(1+5λ) \hat{i}-3λ\hat{j}+(7-3λ) \hat{k}$$
c) $$(1+λ) \hat{i}+λ\hat{j}+(8-3λ) \hat{k}$$
d) $$(1+5λ) \hat{i}-3λ\hat{j}+(4-3λ) \hat{k}$$

Explanation: 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.
∴$$\vec{a}=\hat{i}+4\hat{k}$$
$$\vec{b}=6\hat{i}-3\hat{j}+\hat{k}$$
∴$$\vec{r}=\hat{i}+4\hat{k}+λ(6\hat{i}-3\hat{j}+\hat{k}-(\hat{i}+4\hat{k}))$$
=$$\hat{i}+4\hat{k}+λ(5\hat{i}-3\hat{j}-3\hat{k})$$
=$$(1+5λ) \hat{i}-3λ\hat{j}+(4-3λ) \hat{k}$$

9. Find the vector equation of the line which is passing through the point (1, -4, 4) and parallel to the vector $$2\hat{i}-5\hat{j}+2\hat{k}$$.
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) $$(8+λ) \hat{i}-(4-5λ) \hat{j}+(7-λ) \hat{j}$$

Explanation: We know that, the equation of a vector passing through a point and parallel to another vector is given by $$\vec{r}=\vec{a}+λ\vec{b}$$, where λ is a constant.
The position of vector of the point (1,-4,4) is given by $$\vec{a}=\hat{i}-4\hat{j}+4\hat{k}$$
And $$\vec{b}=2\hat{i}-5\hat{j}+2\hat{k}$$
∴$$\vec{r}=\vec{a}+λ\vec{b}$$
=$$\hat{i}-4\hat{j}+4\hat{k}+λ(2\hat{i}-5\hat{j}+2\hat{k})$$
=$$(1+2λ) \hat{i}-(4+5λ) \hat{j}+(4+2λ) \hat{j}$$

10. Find the cartesian equation of the line which is passing through the point (-1, -8, 5) and parallel to the vector $$7\hat{i}-3\hat{j}-3\hat{k}$$.
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) $$\frac{x+1}{7}=\frac{y+8}{-8}=\frac{z+5}{3}$$

Explanation: The position vector of the given point is $$\vec{a}=-\hat{i}-8\hat{j}+5\hat{k}$$
The vector which is parallel to the given line is $$7\hat{i}-3\hat{j}-3\hat{k}$$
We know that, $$\vec{r}=\vec{a}+λ\vec{b}$$
∴$$x\hat{i}+y\hat{j}+z\hat{k}=-\hat{i}-8\hat{j}+5\hat{k}+λ(7\hat{i}-3\hat{j}-3\hat{k})$$
=$$(-1+7λ) \hat{i}+(-8-3λ) \hat{j}+(5-3λ) \hat{j}$$
⇒$$\frac{x+1}{7}=\frac{y+8}{-3}=\frac{z-5}{-3}$$.

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