Show that the lines through the points(1,-1,2)(3,4,-2)is perpendicular to the line through the points(0,3,2)and(3,5,6).
Solution and Explanation
Let AB be the lines joining the points (1,-1,2) and (3,4,-2), and CD be the lines joining the points, (0,3,2), (3,5,6).
The direction ratios: a1,b1,c1 of AB are (3-1), (4-(-1)), and (-2-2) i.e., 2, 5, and -4.
The direction ratios: a2,b2,c2 of CD are (3-0), (5-3), and (6-2) i.e., 3, 2, and 4.
AB and CD will be perpendicular to each other if a1a2+b1b2+c1c2=0
a1a2+b1b2+c1c2
=2×3+5×2+(-4)×4
=6+10-16
=0
Therefore, AB and CD are perpendicular to each other.
Top Questions on Three Dimensional Geometry
- If $(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from a point $(-1,2,-1)$ to the line joining the points $(2,-1,1)$ and $(1,1,-2)$, then $\alpha+\beta+\gamma=$
- TS EAMCET - 2025
- Mathematics
- Three Dimensional Geometry
- If m:n is the ratio in which the point $\left(\frac{8}{5}, \frac{1}{5}, \frac{8}{5}\right)$ divides the line segment joining the points (2,p,2) and (p,-2,p) where p is an integer then $\frac{3m+n}{3n}=$
- TS EAMCET - 2025
- Mathematics
- Three Dimensional Geometry
- If A(2,1,-1), B(6,-3,2), C(-3,12,4) are the vertices of a triangle ABC and the equation of the plane containing the triangle ABC is $53x+by+cz+d=0$, then $\frac{d}{b+c}=$
- TS EAMCET - 2025
- Mathematics
- Three Dimensional Geometry
- Let $\pi_1$ be the plane determined by the vectors $\hat{i}+\hat{j}, \hat{i}+\hat{k}$ and $\pi_2$ be the plane determined by the vectors $\hat{j}-\hat{k}, \hat{k}-\hat{i}$. Let $\vec{a}$ be a non-zero vector parallel to the line of intersection of the planes $\pi_1$ and $\pi_2$. If $\vec{b} = \hat{i}+\hat{j}-\hat{k}$ then the angle between the vectors $\vec{a}$ and $\vec{b}$ is
- TS EAMCET - 2025
- Mathematics
- Three Dimensional Geometry
Show that the following lines intersect. Also, find their point of intersection:
Line 1: \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \]
Line 2: \[ \frac{x - 4}{5} = \frac{y - 1}{2} = z \]
- CBSE CLASS XII - 2025
- CBSE Compartment XII - 2025
- Mathematics
- Three Dimensional Geometry
Questions Asked in CBSE CLASS XII exam
- The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:
- CBSE CLASS XII - 2025
- Electrostatics
- If \( \int \frac{1}{2x^2} \, dx = k \cdot 2x + C \), then \( k \) is equal to:
- CBSE CLASS XII - 2025
- Integration
- If \( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = 0 \), \( |\overrightarrow{a}| = \sqrt{37} \), \( |\overrightarrow{b}| = 3 \), and \( |\overrightarrow{c}| = 4 \), then the angle between \( \overrightarrow{b} \) and \( \overrightarrow{c} \) is:
- Altima Ltd. invited applications for 2,00,000 equity shares of ₹ 10 at a premium of ₹ 4 per share. Amount payable:
On application and allotment – ₹ 7 (incl. ₹ 1 premium)
On first and final call – Balance.
Applications received for 2,40,000 shares. 30,000 rejected. Manvi allotted 4,000 shares failed to pay first and final call. Her shares were forfeited. These were reissued at ₹ 4 per share fully paid-up.
Pass journal entries in the books of Altima Ltd.- CBSE CLASS XII - 2025
- Accounting for Share Capital
- Determine the values of $x$ for which the function $f(x) = \frac{x-4}{x+1}$ is an increasing or a decreasing function.
- CBSE CLASS XII - 2025
- Application of derivatives
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\(\hat{k}\)