A a - B b - C c are the vertices of a triangle ABC and R r is any point in the plane of triangle ABC- then r - a - b - b - c - c - a is alwaysequal toA- ZeroB- - a b c -C- -a-b-c-D- None of these

The correct option is B [a⃗ b⃗ c⃗] A vector perpendicular to the plane of A (a⃗), B (b⃗), C(c⃗) is (b⃗ a⃗) × (c⃗ a⃗) =a⃗×b⃗+b⃗×c⃗+c⃗×a⃗ Now for any poin ...

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

Byju's Answer

Standard X

Mathematics

Test for Coplanarity

A a , B b , C...

Question

$A (\to a), B (\to b), C (\to c)$ are the vertices of a triangle ABC and $R (\to r)$ is any point in the plane of triangle ABC, then $\to r . (\to a \times \to b + \to b \times \to c + \to c \times \to a)$ is always equal to

Zero

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

$[\to a \to b \to c]$

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

$- [\to a \to b \to c]$

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

None of these

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

Open in App

Solution

The correct option is B
$[\to a \to b \to c]$

A vector perpendicular to the plane of $A (\to a), B (\to b), C (\to c) i s (\to b - \to a) \times (\to c - \to a) = \to a \times \to b + \to b \times \to c + \to c \times \to a$
Now for any point $R (\to r)$ in the plane of A, B, C
$(\to r - \to a) . (\to a \times \to b + \to b \times \to c + \to c \times \to a) = 0 \Rightarrow \to r . (\to a \times \to b + \to b \times \to c + \to c \times \to a) - \to a . (\to a \times \to b + \to b \times \to c + \to c \times a) = 0 \Rightarrow \to r . (\to a \times \to b + \to b \times \to c + \to c \times \to a) = 0 + \to a . \to b \times \to c + 0 \Rightarrow \to r . (\to a \times \to b + \to b \times \to c + \to c \times \to a) = [\to a \to b \to c]$

Suggest Corrections

Similar questions

Q. If

a, b, c

are sides of the

△ A B C

such that

{(1 + \frac{b - c}{a})}^{a} \cdot {(1 + \frac{c - a}{b})}^{b} \cdot {(1 + \frac{a - b}{c})}^{c} \geq 1

, then triangle

△ A B C

must be

Q. Consider the points

A (a, b + c)

B (b, c + a)

, and

C (c, a + b)

be the vertices of

△

ABC. The area of

△

ABC is:

$A (\to a), B (\to b), C (\to c)$ are the vertices of a triangle ABC and $R (\to r)$ is any point in the plane of triangle ABC, then $\to r . (\to a \times \to b + \to b \times \to c + \to c \times \to a)$ is always equal to

The mid points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to

Q. In a triangle

A B C; \frac{a (a + c - b)}{b (b + c - a)}

is equal to

Join BYJU'S Learning Program

A(→a), B(→b), C(→c) are the vertices of a triangle ABC and R(→r) is any point in the plane of triangle ABC, then →r.(→a×→b+→b×→c+→c×→a) is always equal to

$A (\to a), B (\to b), C (\to c)$ are the vertices of a triangle ABC and $R (\to r)$ is any point in the plane of triangle ABC, then $\to r . (\to a \times \to b + \to b \times \to c + \to c \times \to a)$ is always equal to