The required line passes through the origin.
Therefore, its position vector is given by, \(\vec a\)=\(\vec 0\)…………....(1)
The direction ratios of the line through the origin and (5,-2,3) are (5-0)=5, (-2-0)=-2, (3-0)=3
The lines are parallel to the vector given by the equation, b→5\(\hat i\)-2\(\hat j\)+3\(\hat k\)
The equation of the line in vector form through a point with position vector a and parallel to \(\vec b\) is,
\(\vec r = \vec a+\lambda \vec b\), λ∈R
⇒ \(\vec r=\vec 0+\lambda(5\vec i-2\vec j+3\vec k)\)
⇒ \(\vec r=\lambda(5\vec i-2\vec j+3\vec k)\)
The equation of the line through the point (x1,y1,z1) and 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 in the cartesian form is
\(\frac{x-0}{5}=\frac{y-0}{-2}=\frac{z-0}{3}\)
⇒\(\frac{x}{5}=\frac{y}{-2}=\frac{z}{3}\)
List - I | List - II | ||
(P) | γ equals | (1) | \(-\hat{i}-\hat{j}+\hat{k}\) |
(Q) | A possible choice for \(\hat{n}\) is | (2) | \(\sqrt{\frac{3}{2}}\) |
(R) | \(\overrightarrow{OR_1}\) equals | (3) | 1 |
(S) | A possible value of \(\overrightarrow{OR_1}.\hat{n}\) is | (4) | \(\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}\) |
(5) | \(\sqrt{\frac{2}{3}}\) |
A compound (A) with molecular formula $C_4H_9I$ which is a primary alkyl halide, reacts with alcoholic KOH to give compound (B). Compound (B) reacts with HI to give (C) which is an isomer of (A). When (A) reacts with Na metal in the presence of dry ether, it gives a compound (D), C8H18, which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of A, (B), (C) and (D) when (A) reacts with alcoholic KOH.
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.
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\(\hat{k}\)