Find the point of intersection of the following pair of lines, assuming that the vectors $\to a$ and $\to b$ are not parallel.
$\to r = γ (\to b + \to a), \to r = μ (\to b - \to a)$

origin

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\to b + \to a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 \to b

No worries! We‘ve got your back. Try BYJU‘S free classes today!

no intersection point

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B origin
Given,
$\to r = γ (\to b + \to a)$
$\to r = μ (\to b - \to a)$
Equate both equations and find $γ, μ$
$γ (\to b + \to a) = μ (\to b - \to a)$
On comparing we get,
$\Rightarrow (γ - μ) \to b + (γ + μ) \to a = 0$
$\Rightarrow γ - μ = 0, γ + μ = 0$
$\Rightarrow γ = μ = 0$
As $γ = μ = 0$ , So the point of intersection is the origin.

Suggest Corrections

Similar questions

\to a

\to b

\to r = \to a + μ \to b

\to r = \to b + γ \to a

\to a + \to b + \to c = 0, | \to a | = 3, ∣ ∣ \to b ∣ ∣ = 5, | \to c | = 7

\to a

\to b

The point of intersection of the lines $\to r \times \to a = \to b \times \to a$ and $\to r \times \to b = \to a \times \to b$ is

\to a

\to b

(\to a + \to b) \times (\to a - \to b)

\to A

\to B

∣ ∣ \to A + \to B ∣ ∣ = ∣ ∣ \to A - \to B ∣ ∣

Join BYJU'S Learning Program