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Standard XII

Mathematics

Section Formula

If a , b , ...

Question

If $¯ ¯ ¯ a, ¯ ¯ b, ¯ ¯ c$ are the position vectors of the points $A, B, C$ respectively and $3 ¯ ¯ ¯ a + 4 ¯ ¯ b - 7 ¯ ¯ c = ¯ ¯ ¯ 0$ , find the ratio in which point $B$ divides the segment $A C$ .

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Solution

We have
$3 \to a + 4 \to b - 7 \to c = \to 0$
$\Rightarrow 4 \to b = - 3 \to a + 7 \to c$
$\Rightarrow \to b = \frac{- 3 \to a + 7 \to c}{4}$
$\Rightarrow \to b = \frac{- 3 \to a + 7 \to c}{7 - 3}$
$\Rightarrow$ $B$ divides the line segment $A C$ in the ratio $7 : 3$ externally.

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Similar questions

Q. If

¯ ¯ ¯ a, ¯ ¯ b, ¯ ¯ c

are the position vectors of the points

A, B, C

respectively and

2 ¯ ¯ ¯ a + 3 ¯ ¯ b - 5 ¯ ¯ c = ¯ ¯ ¯ 0

, then find the ratio in which the point

C

divides line segment

A B

Q. Let the position vectors of point A, B & C are

\to a

\to b

\to c

respectively. Point D divides line segment BC internally in the ratio 2:1 . Find vector

- - \to A D

Let position vector of the points A,B and C are $\to a, \to b$ and $\to c$ respectively. Point D divides line segment BC internally in the ratio $2 : 1$ . Find vector AD

Q. Show that the four points A, B, C, D with position vectors

\vec{a,} \vec{b,} \vec{c,} \vec{d}

respectively such that

3 \vec{a} - 2 \vec{b} + 5 \vec{c} - 6 \vec{d} = 0,

are coplanar. Also, find the position vector of the point of intersection of the line segments AC and BD.

P

and

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are two points with position vectors

3 \to a - 2 \to b

and

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respectively. Write the position vector of a point R which divides the line segment

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If ¯¯¯a,¯¯b,¯¯c are the position vectors of the points A,B,C respectively and 3¯¯¯a+4¯¯b−7¯¯c=¯¯¯0, find the ratio in which point B divides the segment AC.

We have3→a+4→b−7→c=→0⇒4→b=−3→a+7→c⇒→b=−3→a+7→c4⇒→b=−3→a+7→c7−3⇒ B divides the line segment AC in the ratio 7:3 externally.

If $¯ ¯ ¯ a, ¯ ¯ b, ¯ ¯ c$ are the position vectors of the points $A, B, C$ respectively and $3 ¯ ¯ ¯ a + 4 ¯ ¯ b - 7 ¯ ¯ c = ¯ ¯ ¯ 0$ , find the ratio in which point $B$ divides the segment $A C$ .

We have
$3 \to a + 4 \to b - 7 \to c = \to 0$
$\Rightarrow 4 \to b = - 3 \to a + 7 \to c$
$\Rightarrow \to b = \frac{- 3 \to a + 7 \to c}{4}$
$\Rightarrow \to b = \frac{- 3 \to a + 7 \to c}{7 - 3}$
$\Rightarrow$ $B$ divides the line segment $A C$ in the ratio $7 : 3$ externally.