Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2^i -^j + 4^k$ and is in the direction $^i + 2^j -^k$ .

Open in App

Solution

It is given that the line passes through the point with position vector,
$\to a = 2^i -^j + 4^k . . . (1)$
$\to b =^i + 2^j -^k . . . (2)$
It is known that a line through a point with position vector $\to a$ and parallel to $\to b$ is given by the equation, $\to r = \to a + λ \to b$
$\Rightarrow$ $\to r = 2^i -^j + 4^k + λ (^i + 2^j -^k)$
This is the required equation of the line in vector form.
$\Rightarrow x^i - y^j + z^k = (λ + 2)^i + (2 λ - 1)^j + (- λ + 4)^k$
Eliminating $λ$ , we obtain the Cartesian form equation as
$\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{- 1}$

Suggest Corrections

Similar questions

2^i -^j + 4^k

^i + 2^j -^k

^i +^j - 2^k, 2^i -^j +^k

^i + 2^j +^k

\to r = 3^i -^j -^k + λ (2^i - 2^j +^k)

2^i -^j +^k

-^i + 4^j +^k

^i + 2^j +^k

3^i - 4^j + 5^k

^i + 2^j - 3^k

\to r = (^i +^j -^k) + λ (2^i - 2^j +^k)

\to r = (2^i -^j - 3^k) + μ (^i + 2^j + 2^k)

Join BYJU'S Learning Program