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) \((7+λ) \hat{i}+(4λ+3) \hat{j}+(5-2λ)\hat{k}\)
View Answer

Answer: c
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.
∴\(\vec{r}=2\hat{i}-3\hat{j}+5\hat{k}+λ(3\hat{i}+4\hat{j}-2\hat{k})\)
=\((2+3λ) \hat{i}+(4λ-3) \hat{j}+(5-2λ)\hat{k}\)
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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) \(\frac{x+x_1}{l}=\frac{y+y_1}{m}=\frac{z+z_1}{n}\)
View Answer

Answer: a
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) \(\vec{r}=\vec{a}+λ(\vec{b}-\vec{a})\)
View Answer

Answer: d
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
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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) \((-5+9λ) \hat{i}+(3-6λ)\hat{j}+(1+λ) \hat{k}\)
View Answer

Answer: d
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) \(\frac{2x-1}{3}=\frac{6y+9}{8}=\frac{4z-8}{-2}\)
View Answer

Answer: b
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 \(\frac{x-1}{3}=\frac{y+9}{8}=\frac{z-8}{-2}\)
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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) \(\frac{x-5}{5}=\frac{y+6}{2}=\frac{z-1}{-3}\)
View Answer

Answer: d
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) \(\frac{x-8}{-2}=\frac{y+5}{5}=\frac{z-7}{-7}\)
View Answer

Answer: c
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 \(\frac{x-8}{-1}=\frac{y+5}{6}=\frac{z-7}{-3}\)
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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) \((1+5λ) \hat{i}-3λ\hat{j}+(4-3λ) \hat{k}\)
View Answer

Answer: d
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) \((8+λ) \hat{i}-(4-5λ) \hat{j}+(7-λ) \hat{j}\)
View Answer

Answer: a
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 \(\vec{b}=2\hat{i}-5\hat{j}+2\hat{k}\)
∴\(\vec{r}=\vec{a}+λ\vec{b}\)
=\(\hat{i}-4\hat{j}+4\hat{k}+λ(2\hat{i}-5\hat{j}+2\hat{k})\)
=\((1+2λ) \hat{i}-(4+5λ) \hat{j}+(4+2λ) \hat{j}\)
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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) \(\frac{x+1}{7}=\frac{y+8}{-8}=\frac{z+5}{3}\)
View Answer

Answer: b
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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