MyQuestionIcon

1

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

Ratios of Distances between Centroid, Circumcenter, Incenter and Orthocenter of Triangle

If a⃗, b⃗ and...

Question

If $\to a, \to b a n d \to c$ determine the vertices of a triangle, show that $\frac{1}{2} [\to b \times \to c + \to c \times \to a + \to a \times \to b]$ gives the vector area of the triangle. Hence, deduce the condition that the three points $\to a, \to b a n d \to c$ are collinear. Also, find the vector normal to the plane of the triangle.

Open in App

Solution

Since, $\to a, \to b a n d \to c$ are the vertices of a $△$ ABC as shown.

$Area of △ A B C = \frac{1}{2} | \to A B \times \to A C | N o w, \to A B = \to b - \to a and \to A C = \to c - \to a ∴ Area of △ A B C = \frac{1}{2} | \to b - \to a \times \to c - \to a | = \frac{1}{2} | \to b \times \to c - \to b \times \to a - \to a \times \to c + \to a \times \to a | = \frac{1}{2} | \to b \times \to c + \to a \times \to b + \to c \times \to a + \to 0 | = \frac{1}{2} | \to b \times \to c + \to a \times \to b + \to c \times \to a | . . . (i)$

For three points to be collinear, area of the $△$ ABC should be equal to zero.

$\Rightarrow \frac{1}{2} [\to b \times \to c + \to c \times \to a + \to a \times \to b] = 0 \Rightarrow \to b \times \to c + \to c \times \to a + \to a \times \to b = 0 (i i)$

This is the required condition for collinearity of three points $\to a, \to b a n d \to c$

Let $^n$ be the unit vector normal to the plane of the $△$ ABC.

$∴^n = \frac{\to A B \times \to A C}{| \to A B \times \to A C |} = \frac{\to a \times \to b + \to b \times \to c + \to c \times \to a}{| \to a \times \to b + \to b \times \to c + \to a \times \to a |}$

Suggest Corrections

thumbs-up

6

Similar questions

\to a, \to b, \to c

are three vectors such that

\to a, \to b, \to c

, then prove that

\to a \times \to b = \to b \times \to c = \to c \times \to a

, and hence show that

[\to a \to b \to c] = 0.

\to a, \to b, \to c

are three vectors such that

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

then prove that

\to a \times \to b = \to b \times \to c = \to c \times \to a,

and hence show that

[\to a \to b \to c] = 0.

Q. For any three vectors

\to a, \to b

\to c

[\to a + \to b, \to b + \to c, \to c + \to a] = 2 [\to a \to b \to c]

. Hence prove that the vectors

\to a + \to b, \to b + \to c, \to c + \to a

are coplanar. If and only if

\to a, \to b, \to c

\to a, \to b, \to c

are three vectors such that

\to a \times \to b = \to c, \to b \times \to c = \to a, \to c \times \to a = \to b

then prove that

| \to a | = | \to b | = | \to c |

\to a, \to b, \to c

are the vertices of the triangle

A B C

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

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Ratios of Distances between Centroid, Circumcenter, Incenter and Orthocenter of Triangle

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon