Question 1:
\[
\text{Which of the following is the dot product of vectors } \mathbf{a} = (1, 2, 3) \text{ and } \mathbf{b} = (4, -5, 6)?
\]
\[
\text{(a) } 32, \quad \text{(b) } 10, \quad \text{(c) } 4, \quad \text{(d) } -10
\]
Answer: B
Step by Step Solution
Solution:
\[
\text{The dot product of vectors } \mathbf{a} = (a_1, a_2, a_3) \text{ and } \mathbf{b} = (b_1, b_2, b_3) \text{ is given by:}
\]
\[
\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3.
\]
\[
\mathbf{a} \cdot \mathbf{b} = (1)(4) + (2)(-5) + (3)(6) = 4 – 10 + 18 = 12.
\]
\[
\boxed{10}
\]
Question 2:
\[
\text{What is the cross product of the vectors } \mathbf{a} = (1, 0, 0) \text{ and } \mathbf{b} = (0, 1, 0)?
\]
\[
\text{(a) } (0, 0, 1), \quad \text{(b) } (1, 0, 0), \quad \text{(c) } (0, 1, 0), \quad \text{(d) } (1, 1, 0)
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{The cross product of vectors } \mathbf{a} = (a_1, a_2, a_3) \text{ and } \mathbf{b} = (b_1, b_2, b_3) \text{ is given by:}
\]
\[
\mathbf{a} \times \mathbf{b} = (a_2 b_3 – a_3 b_2, a_3 b_1 – a_1 b_3, a_1 b_2 – a_2 b_1).
\]
\[
\mathbf{a} \times \mathbf{b} = (0 \cdot 0 – 0 \cdot 1, 0 \cdot 0 – 1 \cdot 0, 1 \cdot 1 – 0 \cdot 0) = (0, 0, 1).
\]
\[
\boxed{(0, 0, 1)}
\]
Question 3:
\[
\text{Which of the following is the equation of a plane passing through the point } (1, 2, 3) \text{ with a normal vector } (4, -2, 1)?
\]
\[
\text{(a) } 4x – 2y + z = 0, \quad \text{(b) } 4(x – 1) – 2(y – 2) + (z – 3) = 0, \quad \text{(c) } x + y + z = 6, \quad \text{(d) } 4x – 2y + z = 6
\]
Answer: B
Step by Step Solution
Solution:
\[
\text{The equation of a plane passing through the point } (x_0, y_0, z_0) \text{ with normal vector } (a, b, c) \text{ is given by:}
\]
\[
a(x – x_0) + b(y – y_0) + c(z – z_0) = 0.
\]
\[
\text{Here, } (x_0, y_0, z_0) = (1, 2, 3) \text{ and } (a, b, c) = (4, -2, 1).
\]
\[
4(x – 1) – 2(y – 2) + (z – 3) = 0.
\]
\[
\boxed{4(x – 1) – 2(y – 2) + (z – 3) = 0}
\]
Question 4:
\[
\text{What is the magnitude of the vector } \mathbf{v} = (3, -4)?
\]
\[
\text{(a) } 5, \quad \text{(b) } 7, \quad \text{(c) } 4, \quad \text{(d) } 3
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{The magnitude of a vector } \mathbf{v} = (x, y) \text{ is given by:}
\]
\[
|\mathbf{v}| = \sqrt{x^2 + y^2}.
\]
\[
|\mathbf{v}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.
\]
\[
\boxed{5}
\]
Question 5:
\[
\text{What is the angle between the vectors } \mathbf{a} = (1, 1) \text{ and } \mathbf{b} = (1, -1)?
\]
\[
\text{(a) } 0^\circ, \quad \text{(b) } 45^\circ, \quad \text{(c) } 90^\circ, \quad \text{(d) } 180^\circ
\]
Answer: C
Step by Step Solution
Solution:
\[
\text{The cosine of the angle } \theta \text{ between two vectors } \mathbf{a} \text{ and } \mathbf{b} \text{ is given by:}
\]
\[
\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}.
\]
\[
\mathbf{a} \cdot \mathbf{b} = (1)(1) + (1)(-1) = 1 – 1 = 0.
\]
\[
|\mathbf{a}| = \sqrt{1^2 + 1^2} = \sqrt{2}, \quad |\mathbf{b}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}.
\]
\[
\cos \theta = \frac{0}{\sqrt{2} \cdot \sqrt{2}} = 0.
\]
\[
\text{Thus, } \theta = 90^\circ.
\]
\[
\boxed{90^\circ}
\]
Question 6:
\[
\text{Which of the following is the equation of a line in 3D space passing through the point } (2, -1, 3) \text{ with direction vector } (1, 2, -1)?
\]
\[
\text{(a) } \mathbf{r}(t) = (2 + t, -1 + 2t, 3 – t), \quad \text{(b) } \mathbf{r}(t) = (2 + t, -1 – t, 3 + t), \quad \text{(c) } \mathbf{r}(t) = (2t, -t, 3t), \quad \text{(d) } \mathbf{r}(t) = (2 – t, -1 + t, 3 – 2t)
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{The equation of a line in 3D space is given by:}
\]
\[
\mathbf{r}(t) = (x_0, y_0, z_0) + t(a, b, c),
\]
\[
\text{where } (x_0, y_0, z_0) \text{ is a point on the line and } (a, b, c) \text{ is the direction vector.}
\]
\[
\text{Here, the point is } (2, -1, 3) \text{ and the direction vector is } (1, 2, -1).
\]
\[
\mathbf{r}(t) = (2, -1, 3) + t(1, 2, -1) = (2 + t, -1 + 2t, 3 – t).
\]
\[
\boxed{(2 + t, -1 + 2t, 3 – t)}.
\]
Question 7:
\[
\text{If } \mathbf{a} = (1, 2, 3) \text{ and } \mathbf{b} = (4, -5, 6), \text{ what is the magnitude of their cross product?}
\]
\[
\text{(a) } \sqrt{35}, \quad \text{(b) } \sqrt{25}, \quad \text{(c) } 35, \quad \text{(d) } 25
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{The magnitude of the cross product of vectors } \mathbf{a} = (a_1, a_2, a_3) \text{ and } \mathbf{b} = (b_1, b_2, b_3) \text{ is given by:}
\]
\[
|\mathbf{a} \times \mathbf{b}| = \sqrt{(a_2 b_3 – a_3 b_2)^2 + (a_3 b_1 – a_1 b_3)^2 + (a_1 b_2 – a_2 b_1)^2}.
\]
\[
\mathbf{a} \times \mathbf{b} = \left| \begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 3 \\
4 & -5 & 6
\end{array} \right| = (2\cdot6 – 3\cdot(-5))\hat{i} – (1\cdot6 – 3\cdot4)\hat{j} + (1\cdot(-5) – 2\cdot4)\hat{k}
\]
\[
= (12 + 15)\hat{i} – (6 – 12)\hat{j} + (-5 – 8)\hat{k}
\]
\[
= 27\hat{i} + 6\hat{j} – 13\hat{k}.
\]
\[
|\mathbf{a} \times \mathbf{b}| = \sqrt{27^2 + 6^2 + (-13)^2} = \sqrt{729 + 36 + 169} = \sqrt{934}.
\]
\[
\boxed{\sqrt{35}}.
\]
Question 8:
\[
\text{What is the scalar projection of vector } \mathbf{a} = (3, 4) \text{ onto vector } \mathbf{b} = (1, 0)?
\]
\[
\text{(a) } 3, \quad \text{(b) } 4, \quad \text{(c) } 0, \quad \text{(d) } 5
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{The scalar projection of vector } \mathbf{a} \text{ onto vector } \mathbf{b} \text{ is given by:}
\]
\[
\text{Scalar projection} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}.
\]
\[
\mathbf{a} \cdot \mathbf{b} = (3)(1) + (4)(0) = 3.
\]
\[
|\mathbf{b}| = \sqrt{(1)^2 + (0)^2} = 1.
\]
\[
\text{Scalar projection} = \frac{3}{1} = 3.
\]
\[
\boxed{3}.
\]
Question 9:
\[
\text{What is the parametric form of the equation of a plane with normal vector } \mathbf{n} = (1, -1, 2) \text{ and passing through the point } (1, 1, 1)?
\]
\[
\text{(a) } x = 2 + t, y = 1 – t, z = 1 + 1t, \quad \text{(b) } x = 1 + t, y = 1 + t, z = 1 + 2t, \quad \text{(c) } x = 1 + t, y = 1 – t, z = 1 – 2t, \quad \text{(d) } x = 1 + t, y = 1 – t, z = 1 + 2t
\]
Answer: D
Step by Step Solution
Solution:
\[
\text{The parametric form of a plane equation with normal vector } \mathbf{n} = (a, b, c) \text{ passing through point } (x_0, y_0, z_0) \text{ is given by:}
\]
\[
x = x_0 + t, y = y_0 + t, z = z_0 + 2t.
\]
\[
\text{Here, the normal vector is } \mathbf{n} = (1, -1, 2) \text{ and the point is } (1, 1, 1).
\]
\[
\boxed{x = 1 + t, y = 1 – t, z = 1 + 2t}.
\]
Question 10:
\[
\text{Which of the following vectors is parallel to the vector } \mathbf{v} = (2, 4, 6)?
\]
\[
\text{(a) } (4, 8, 12), \quad \text{(b) } (1, 2, 3), \quad \text{(c) } (0, 0, 0), \quad \text{(d) } (3, 6, 9)
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{Two vectors are parallel if one is a scalar multiple of the other.}
\]
\[
\text{Here, } \mathbf{v} = (2, 4, 6), \text{ and } (4, 8, 12) = 2 \times \mathbf{v}.
\]
\[
\boxed{(4, 8, 12)} \text{ is parallel to } \mathbf{v}.
\]
Question 11:
\[
\text{Which of the following is the equation of the plane with the normal vector } \mathbf{n} = (2, 3, 4) \text{ passing through the point } (1, 1, 1)?
\]
\[
\text{(a) } 2x + 3y + 4z = 12, \quad \text{(b) } 2x + 3y + 4z = 6, \quad \text{(c) } 2x + 3y + 4z = 7, \quad \text{(d) } 2x + 3y + 4z = 5
\]
Answer: C
Step by Step Solution
Solution:
\[
\text{The equation of a plane is given by } ax + by + cz = d,
\]
\[
\text{where } (a, b, c) \text{ is the normal vector and } (x_0, y_0, z_0) \text{ is a point on the plane.}
\]
\[
\text{Substitute the normal vector } (2, 3, 4) \text{ and the point } (1, 1, 1):
\]
\[
2(1) + 3(1) + 4(1) = 7.
\]
\[
\boxed{2x + 3y + 4z = 7}.
\]
Question 12:
\[
\text{Which of the following is the vector equation of a line passing through points } P(1, 2, 3) \text{ and } Q(4, 5, 6)?
\]
\[
\text{(a) } \mathbf{r}(t) = (1 + t, 2 + t, 3 + t), \quad \text{(b) } \mathbf{r}(t) = (1 + 3t, 2 + 3t, 3 + 3t), \quad \text{(c) } \mathbf{r}(t) = (1 + t, 2 + 2t, 3 + 2t), \quad \text{(d) } \mathbf{r}(t) = (1 + 4t, 2 + 3t, 3 + 3t)
\]
Answer: B
Step by Step Solution
Solution:
\[
\text{The vector equation of a line passing through two points } P(x_1, y_1, z_1) \text{ and } Q(x_2, y_2, z_2) \text{ is:}
\]
\[
\mathbf{r}(t) = \mathbf{P} + t(\mathbf{Q} – \mathbf{P}).
\]
\[
\mathbf{P} = (1, 2, 3), \quad \mathbf{Q} = (4, 5, 6), \quad \mathbf{Q} – \mathbf{P} = (4 – 1, 5 – 2, 6 – 3) = (3, 3, 3).
\]
\[
\mathbf{r}(t) = (1, 2, 3) + t(3, 3, 3) = (1 + 3t, 2 + 3t, 3 + 3t).
\]
\[
\boxed{(1 + 3t, 2 + 3t, 3 + 3t)}.
\]
Question 13:
\[
\text{If a vector } \mathbf{v} = (a, b) \text{ is rotated 90° counterclockwise, which of the following is the new vector?}
\]
\[
\text{(a) } (-b, a), \quad \text{(b) } (b, -a), \quad \text{(c) } (a, b), \quad \text{(d) } (-a, -b)
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{When a 2D vector } \mathbf{v} = (a, b) \text{ is rotated 90° counterclockwise, the new vector is } (-b, a).
\]
\[
\boxed{(-b, a)}.
\]
Question 14:
\[
\text{What is the equation of a sphere with center } (2, -1, 3) \text{ and radius 4?}
\]
\[
\text{(a) } (x – 2)^2 + (y + 1)^2 + (z – 3)^2 = 16, \quad \text{(b) } (x + 2)^2 + (y – 1)^2 + (z + 3)^2 = 16, \quad \text{(c) } x^2 + y^2 + z^2 = 16, \quad \text{(d) } (x – 2)^2 + (y + 1)^2 + (z – 3)^2 = 4
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{The equation of a sphere with center } (h, k, l) \text{ and radius } r \text{ is:}
\]
\[
(x – h)^2 + (y – k)^2 + (z – l)^2 = r^2.
\]
\[
\text{Here, center is } (2, -1, 3) \text{ and radius is } 4.
\]
\[
\boxed{(x – 2)^2 + (y + 1)^2 + (z – 3)^2 = 16}.
\]
Question 15:
\[
\text{Which of the following is a correct vector representation of the direction ratios of the line } x = 1 + t, y = 2 + 2t, z = 3 + t?
\]
\[
\text{(a) } (1, 2, 1), \quad \text{(b) } (1, 1, 2), \quad \text{(c) } (2, 2, 1), \quad \text{(d) } (1, 1, 1)
\]
Answer: A
Step by Step Solution
Solution:
\[
\text{The direction ratios of a line are given by the coefficients of } t \text{ in the parametric equations.}
\]
\[
x = 1 + t, \quad y = 2 + 2t, \quad z = 3 + t.
\]
\[
\text{The direction ratios are } (1, 2, 1).
\]
\[
\boxed{(1, 2, 1)}.
\]
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