To find the equation of the line passing through the points \((-3, 4, 11)\) and \((1, -2, 7)\), we use the vector form of a line in three-dimensional space. The equation for a line passing through the point \((x_1, y_1, z_1)\) with direction ratios \((a, b, c)\) is given by:
\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\)
First, let's determine the direction ratios by subtracting the coordinates of the given points:
Thus, the direction ratios are \((4, -6, -4)\). However, these ratios can be simplified by dividing each by 2:
Resulting simplified direction ratios are \((2, -3, -2)\).
Now, substitute these values into the line equation form using the point \((-3, 4, 11)\):
\(\frac{x+3}{2}=\frac{y-4}{-3}=\frac{z-11}{-2}\)
To compare with the given options, we observe that by multiplying each term by -1, we get the correct formation:
\(\frac{x+3}{-2}=\frac{y-4}{3}=\frac{z-11}{2}\)
This matches the given correct option: \(\frac{x+3}{-2}=\frac{y-4}{3}=\frac{z-11}{2}\).
Therefore, the correct option is:
\(\frac{x+3}{-2}=\frac{y-4}{3}=\frac{z-11}{2}\)
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}
Match the following:
In the following, \( [x] \) denotes the greatest integer less than or equal to \( x \). 
Choose the correct answer from the options given below:
For x < 0:
f(x) = ex + ax
For x ≥ 0:
f(x) = b(x - 1)2
Mathematically, Geometry is one of the most important topics. The concepts of Geometry are derived w.r.t. the planes. So, Geometry is divided into three major categories based on its dimensions which are one-dimensional geometry, two-dimensional geometry, and three-dimensional geometry.
Consider a line L that is passing through the three-dimensional plane. Now, x,y and z are the axes of the plane and α,β, and γ are the three angles the line makes with these axes. These are commonly known as the direction angles of the plane. So, appropriately, we can say that cosα, cosβ, and cosγ are the direction cosines of the given line L.
