CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Dot Product

If a⃗, b⃗, ...

Question

If $\to a, \to b, \to c$ are the position vectors of the vertices of an equilateral triangle whose orthocentre is at the origin, then

A

\to a + \to b + \to c = \to 0

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

B

{\to a}^{2} = {\to b}^{2} + {\to c}^{2}

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

C

\to a + \to b = \to c

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

D

None of these

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

Open in App

Solution

The correct option is A $\to a + \to b + \to c = \to 0$
$\to a, \to b, \to c$ are the position vectors of the vertices of an equilateral triangle
Then centroid of the triangle is $\frac{\to a + \to b + \to c}{3}$ .
Since, the triangle is an equilateral, therefore the orthocentre coincides with the centroid and hence
$\frac{\to a + \to b + \to c}{3} = \to 0 \Rightarrow \to a + \to b + \to c = \to 0$ .

Suggest Corrections

thumbs-up

0

Similar questions

\to a, \to b, \to c

are unit vectors such that

\to a + \to b + \to c = \to 0

(\to a, \to b) = \frac{π}{3}

| \to a \times \to b | + | \to b \times \to c | + | \to c \times \to a | =

\to a, \to b, \to c

are unit vectors such that

\to a + \to b + \to c = \to 0

(\to a, \to b) = \frac{π}{3}

∣ ∣ \to a \times \to b ∣ ∣ + ∣ ∣ \to b \times \to c ∣ ∣ + | \to c \times \to a | =

\to a, \to b, \to c

are positive vectors of vertices

A, B, C

Δ A B C

\to r

is position vector of a point

P

(| \to b - \to c | + | \to c - \to a | + | \to a - \to b |) \to r = | \to b - \to c | \to a + | \to c - \to a | \to b + ∣ ∣ \to a - \to b ∣ ∣ \to c

P

\to a, \to b, \to c

are three non-zero, non-coplanar vectors and

{\to b}_{1} = \to b - \frac{\to b \cdot \to a}{| \to a |^{2}} \to a,

{\to b}_{2} = \to b + \frac{\to b \cdot \to a}{| \to a |^{2}} \to a

{\to c}_{1} = \to c - \frac{\to c \cdot \to a}{| \to a |^{2}} \to a + \frac{\to b \cdot \to c}{| \to c |^{2}} {\to b}_{1},

{\to c}_{2} = \to c - \frac{\to c \cdot \to a}{| \to a |^{2}} \to a - \frac{{\to b}_{1} \cdot \to c}{| {\to b}_{1} |^{2}} {\to b}_{1},

{\to c}_{3} = \to c - \frac{\to c \cdot \to a}{| \to c |^{2}} \to a + \frac{\to b \cdot \to c}{| \to c |^{2}} {\to b}_{1}, {\to c}_{4} = \to c - \frac{\to c \cdot \to a}{| \to c |^{2}} \to a - \frac{\to b \cdot \to c}{| \to b |^{2}} {\to b}_{1}

, then the set of orthogonal vectors is

Q. Prove that points

A, B, C

having positions vectors

\to a, \to b, \to c

are collinear, if

[\to b \times \to c + \to c \times \to a + \to a \times \to b] = \to 0

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon