The cartesian equation of a line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\). Write its vector form.
The cartesian equation of a line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\). Write its vector form.
Solution and Explanation
The Cartesian equation of the line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\)……….....(1)
The given line passes through the point (5,-4,6).
The position vector of this point is \(\vec a = 5\hat i-4\hat j +6 \hat k\)
Also, the direction ratios of the given line are 3, 7, and 2.
This means that the line is in the direction of the vector, \(\vec b\)=3\(\hat i\)+7\(\hat j\)+2\(\hat k\)
It is known that the line through position vector \(\vec a\) and in the direction of the vector \(\vec b\) is given by the equation,
\(\vec r\)=\(\vec a\)+λ\(\vec b\), λ∈R
⇒ \(\vec r\)=(5\(\vec i\)-4\(\vec j\)+6\(\vec k\))+λ(3\(\vec i\)+7\(\vec j\)+2\(\vec k\))
This is the required equation of the given line in vector form.
Top Questions on Three Dimensional Geometry
- Let \( L \) be the line \[ \frac{x+1}{2} = \frac{y+1}{3} = \frac{z+3}{6} \] and let \( S \) be the set of all points \( (a,b,c) \) on \( L \), whose distance from the line \[ \frac{x+1}{2} = \frac{y+1}{3} = \frac{z-9}{0} \] along the line \( L \) is \( 7 \). Then \[ \sum_{(a,b,c)\in S} (a+b+c) \] is equal to
- JEE Main - 2026
- Mathematics
- Three Dimensional Geometry
- If the distances of the point \( (1,2,a) \) from the line \[ \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} \] along the lines \[ L_1:\ \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b} \quad \text{and} \quad L_2:\ \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c} \] are equal, then \( a+b+c \) is equal to:
- JEE Main - 2026
- Mathematics
- Three Dimensional Geometry
- Let \(L\) be the distance of point \(P(-1,2,5)\) from the line \[ \frac{x-1}{2}=\frac{y-3}{2}=\frac{z+1}{1} \] measured {parallel to a line having direction ratios \(4,\,3,\,-5\). Then \(L^2\) is equal to: }
- JEE Main - 2026
- Mathematics
- Three Dimensional Geometry
- If the image of a point \(P(3,2,1)\) in the line \[ \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} \] is \(Q\), then the distance of \(Q\) from the line \[ \frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2} \] is:
- JEE Main - 2026
- Mathematics
- Three Dimensional Geometry
- The value of the integral \( \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1 + \sqrt[3]{\tan 2x}} \) is :
- JEE Main - 2026
- Mathematics
- Three Dimensional Geometry
Questions Asked in CBSE CLASS XII exam
- Find the general solution of the differential equation \[ y\log y\,\frac{dx}{dy}+x=\frac{2}{y}. \]
- CBSE CLASS XII - 2026
- Differential Equations
- A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:
- CBSE CLASS XII - 2026
- Electromagnetic induction
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following:
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).
- CBSE CLASS XII - 2026
- Integration
- Consider a cylindrical conductor of length \( l \) and area of cross-section \( A \). Current \( I \) is maintained in the conductor and electrons drift with velocity \( \vec{v}_d \, (|\vec{v}_d| = \frac{eE}{m} \tau) \), where symbols have their usual meanings. Show that the conductivity of the material of the conductor is given by \[ \sigma = \frac{n e^2 \tau}{m}. \]
- CBSE CLASS XII - 2026
- Current Density
- The painting Hazrat Nizamuddin Auliya and Amir Khusro belongs to which sub-school of Deccan Miniature?
- CBSE CLASS XII - 2026
- Indian Miniature Paintings and Museums
Concepts Used:
Equation of a Line in Space
In a plane, the equation of a line is given by the popular equation y = m x + C. Let's look at how the equation of a line is written in vector form and Cartesian form.
Vector Equation
Consider a line that passes through a given point, say ‘A’, and the line is parallel to a given vector '\(\vec{b}\)‘. Here, the line ’l' is given to pass through ‘A’, whose position vector is given by '\(\vec{a}\)‘. Now, consider another arbitrary point ’P' on the given line, where the position vector of 'P' is given by '\(\vec{r}\)'.
\(\vec{AP}\)=𝜆\(\vec{b}\)
Also, we can write vector AP in the following manner:
\(\vec{AP}\)=\(\vec{OP}\)–\(\vec{OA}\)
𝜆\(\vec{b}\) =\(\vec{r}\)–\(\vec{a}\)
\(\vec{a}\)=\(\vec{a}\)+𝜆\(\vec{b}\)
\(\vec{b}\)=𝑏1\(\hat{i}\)+𝑏2\(\hat{j}\) +𝑏3\(\hat{k}\)