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Assam CEE Section Formula — practice questions

Explore 4 free Section Formula questions in Assam CEE Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Section Formula questions and exercises with answers

Showing questions 1–4 of 4.

The position vector of the **midpoint** of the segment joining points with position vectors $\vec a$ and $\vec b$ is:Read the answer and worked solution →The point dividing the line segment joining $A(2, -6, 8)$ and $B(-1, 3, -4)$ internally in ratio $1:3$ has coordinates:Read the answer and worked solution →If $\vec a, \vec b, \vec c$ are the position vectors of $A, B, C$ and $5\vec a - 3\vec b - 2\vec c = \vec 0$, then point $C$ divides $\overline{BA}$:Read the answer and worked solution →Points $A(3, 2, p), B(q, 8, -10), C(-2, -3, 1)$ are collinear. The values of $p$ and $q$ are:Read the answer and worked solution →

Worked practice questions

The position vector of the **midpoint** of the segment joining points with position vectors $\vec a$ and $\vec b$ is:

  1. $(\vec a + \vec b)/2$
  2. $(\vec a - \vec b)/2$
  3. $\vec a \cdot \vec b$
  4. $|\vec a| + |\vec b|$
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Correct answer: A. $(\vec a + \vec b)/2$

Midpoint formula: set m = n = 1 in the section formula to get $\vec r = (\vec a + \vec b)/2$.

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The point dividing the line segment joining $A(2, -6, 8)$ and $B(-1, 3, -4)$ internally in ratio $1:3$ has coordinates:

  1. $(5/4, -15/4, 5)$
  2. $(-7/2, 21/2, -14)$
  3. $(1/4, -1/4, 2)$
  4. $(7/2, -21/2, 14)$
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Correct answer: A. $(5/4, -15/4, 5)$

$\vec r = (1\cdot \vec b + 3\cdot \vec a)/(1+3) = (1(-\hat i+3\hat j-4\hat k) + 3(2\hat i-6\hat j+8\hat k))/4 = (5\hat i -15\hat j +20\hat k)/4$ giving $(5/4, -15/4, 5)$.

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If $\vec a, \vec b, \vec c$ are the position vectors of $A, B, C$ and $5\vec a - 3\vec b - 2\vec c = \vec 0$, then point $C$ divides $\overline{BA}$:

  1. Internally in ratio 5:3
  2. Externally in ratio 5:3
  3. Internally in ratio 3:5
  4. Externally in ratio 3:5
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Correct answer: B. Externally in ratio 5:3

**From the relation:** $2\vec c = 5\vec a - 3\vec b$ ⇒ $\vec c = (5\vec a - 3\vec b)/2 = (5\vec a - 3\vec b)/(5 - 3)$. This matches the external division formula $\vec r = (m\vec A - n\vec B)/(m - n)$ with $m = 5, n = 3$ so $C$ divides $\overline{BA}$ externally in ratio 5:3.

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