Find the cartesian equation of the line that passes through the point(-2,4,-5)and parallel to the line given by \(\frac{x+3}{3}\)=\(\frac{y-4}{5}\)=\(\frac{z+8}{6}\)
Find the cartesian equation of the line that passes through the point(-2,4,-5)and parallel to the line given by \(\frac{x+3}{3}\)=\(\frac{y-4}{5}\)=\(\frac{z+8}{6}\)
Solution and Explanation
It is given that the line passes through the point (-2,4,-5) and is parallel to \(\frac{x+3}{3}\)=\(\frac{y-4}{5}\)=\(\frac{z+8}{6}\)
The direction ratios of the line, \(\frac{x+3}{3}\)=\(\frac{y-4}{5}\)=\(\frac{z+8}{6}\), are 3, 5, and 6.
The required line is parallel to \(\frac{x+3}{3}\)=\(\frac{y-4}{5}\)=\(\frac{z+8}{6}\)
Therefore, its direction ratios are 3k, 5k, and 6k, where k≠0
It is known that the equation of the line through the point (x1,y1,z1) and with direction ratios, a,b,c is given by \(\frac{x-x_1}{a}\)=\(\frac{y-y_1}{b}\)=\(\frac{z-z_1}{c}\)
Therefore, the equation of the required line is
\(\Rightarrow \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}=k\)
Top Questions on 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
- 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
- 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
- Emily, Flora and Ginni entered into a partnership on 1st October, 2023 with capitals of ₹ 10,00,000 each. The partnership deed provided for interest on capital at 10% p.a. The firm earned a net profit of ₹ 7,50,000 for the year ended 31st March, 2024. The amount of profit transferred to Emily’s capital account was :
- CBSE CLASS XII - 2025
- Partnership Accounts
- "The Government had launched Incredible India Campaign to promote tourism in various parts of the country."
Elaborate the impact of Incredible India Campaign on foreign exchange reserves and Balance of Payment of India.- CBSE CLASS XII - 2025
- Balance of payments (BOP)
- Ajanta Ltd. purchased machinery worth Rs 36,00,000 from Sujata Ltd. Ajanta Ltd. paid half the amount to Sujata Ltd. through a bank draft and the balance by issuing 8 debentures of Rs 100 each at a discount of 10.
Pass the necessary journal entries in the books of Ajanta Ltd. for the above transactions.- CBSE CLASS XII - 2025
- Debentures
- From the following information obtained from the books of KVK Ltd., calculate Net Assets Turnover Ratio and Debt Equity Ratio:
Information Table
Information Amount (₹) Preference Share Capital 8,00,000 Equity Share Capital 12,00,000 General Reserve 2,00,000 Balance in Statement of Profit and Loss 6,00,000 15% Debentures 4,00,000 12% Loan 4,00,000 Revenue from Operations 72,00,000 - CBSE CLASS XII - 2025
- Financial Statement Analysis
- Zaina, Yash and Kiran were partners in a firm sharing profits and losses in the ratio of 2 : 2 : 1. Zaina died on 1st July, 2024. As per the partnership deed, Zaina’s share of profit or loss till the date of her death was to be calculated on the basis of sales. Sales for the year ended 31st March, 2024 amounted to ₹ 4,00,000 and that from 1st April to 30th June, 2024 was ₹ 1,50,000. The profit for the year ending 31st March, 2024 was ₹ 1,00,000. Calculate Zaina’s share of profit in the firm till the date of her death and pass necessary journal entry for the same.
- CBSE CLASS XII - 2025
- Retirement and Death of a Partner
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\(\hat{k}\)