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GRE Unit 9A: Parametric Equations and Plane Curves — practice questions

Explore 50 free Unit 9A: Parametric Equations and Plane Curves questions in GRE Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Unit 9A: Parametric Equations and Plane Curves questions and exercises with answers

Showing questions 1–30 of 50.

Find the value of $\lambda$ and $\mu$ if the non-zero, non-collinear vectors $\vec a$ and $\vec b$ satisfy $\vec a + 3\vec b = 2\lambda \vec a - \mu \vec b$:Read the answer and worked solution →Vector $4\hat i + 13\hat j - 18\hat k$ expressed as a linear combination of $\hat i - 2\hat j + 3\hat k$ and $2\hat i + 3\hat j - 4\hat k$ has coefficients $(m, n)$: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 →If $G(a, 2, -1)$ is the centroid of the triangle with vertices $P(1, 3, 2),\ Q(3, b, -4),\ R(5, 1, c)$, then $(a, b, c)$ equals:Read the answer and worked solution →The centroid of the tetrahedron with vertices $A(3, -5, 7),\ B(5, 4, 2),\ C(7, -7, -3),\ D(1, 0, 2)$ is:Read the answer and worked solution →If $A(0, 3, 0), B(0, 0, 4), C(0, 3, 4)$ are vertices of a triangle, the position vector of its **incentre** is: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 →A quantity having only magnitude (no direction) is called:Read the answer and worked solution →Vector addition is:Read the answer and worked solution →The dot product ā · b̄ is:Read the answer and worked solution →The cross product ā × b̄ is:Read the answer and worked solution →What is î × ĵ?Read the answer and worked solution →Two vectors are PERPENDICULAR iff:Read the answer and worked solution →Two vectors are PARALLEL (or anti-parallel) iff:Read the answer and worked solution →The MAGNITUDE of a × b equals:Read the answer and worked solution →Find magnitude of ā = 3î + 4ĵ:Read the answer and worked solution →Find ā · b̄ if ā = 2î + 3ĵ and b̄ = î - 2ĵ:Read the answer and worked solution →Find ā × b̄ where ā = î + 2ĵ + k̂ and b̄ = 2î + ĵ + 3k̂:Read the answer and worked solution →Angle between two vectors ā = 2î + 2ĵ and b̄ = 4î - 4ĵ:Read the answer and worked solution →Find unit vector in direction of ā = 3î + 4ĵ:Read the answer and worked solution →Area of parallelogram formed by ā and b̄ equals:Read the answer and worked solution →Position vectors of P and Q are p̄ and q̄. The point R divides PQ internally in ratio m:n. Position vector of R:Read the answer and worked solution →Three vectors ā, b̄, c̄ are coplanar iff their SCALAR TRIPLE PRODUCT [ā b̄ c̄] equals:Read the answer and worked solution →For vectors ā = î + ĵ and b̄ = î - ĵ, find |ā + b̄|:Read the answer and worked solution →Projection of ā = 3î + 4ĵ on b̄ = î + ĵ:Read the answer and worked solution →If ā × b̄ = 0 and neither is the zero vector, then:Read the answer and worked solution →Volume of parallelepiped formed by ā, b̄, c̄:Read the answer and worked solution →For unit vectors î, ĵ, k̂: î · ĵ + ĵ · k̂ + k̂ · î =Read the answer and worked solution →For vectors ā = î - ĵ + 2k̂ and b̄ = 3î + 2ĵ - k̂, find the AREA of the parallelogram formed (use |ā × b̄|):Read the answer and worked solution →Find cos θ for ā = 2î + 3ĵ - 4k̂ and b̄ = î + 2ĵ + 3k̂ (where θ is the angle between them):Read the answer and worked solution →

Worked practice questions

Find the value of $\lambda$ and $\mu$ if the non-zero, non-collinear vectors $\vec a$ and $\vec b$ satisfy $\vec a + 3\vec b = 2\lambda \vec a - \mu \vec b$:

  1. $\lambda = 1/2,\ \mu = -3$
  2. $\lambda = -1/2,\ \mu = 3$
  3. $\lambda = 2,\ \mu = -3$
  4. $\lambda = 1,\ \mu = -3$
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Correct answer: A. $\lambda = 1/2,\ \mu = -3$

Non-collinear $\vec a, \vec b$ form a basis ⇒ coefficients on each side must match. Coefficient of $\vec a$: $1 = 2\lambda$ ⇒ $\lambda = 1/2$. Coefficient of $\vec b$: $3 = -\mu$ ⇒ $\mu = -3$.

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Vector $4\hat i + 13\hat j - 18\hat k$ expressed as a linear combination of $\hat i - 2\hat j + 3\hat k$ and $2\hat i + 3\hat j - 4\hat k$ has coefficients $(m, n)$:

  1. $(2, 3)$
  2. $(-2, 3)$
  3. $(3, -2)$
  4. $(-3, 2)$
Show answer and solution

Correct answer: B. $(-2, 3)$

Set $4\hat i + 13\hat j - 18\hat k = m(\hat i - 2\hat j + 3\hat k) + n(2\hat i + 3\hat j - 4\hat k)$. **Equating components:** $m + 2n = 4$ and $-2m + 3n = 13$. **Solving:** $m = -2, n = 3$. Check the $\hat k$ component: $3(-2) - 4(3) = -18$ ✓.

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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
Show answer and solution

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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More Mathematics topics

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