Question

(1)The scalar product of vectors a and b, where θ is the angle between them, is ......

(2)Determine the gradient of a straight line passing through the point (1, 6) and (-3, -3).

(3)Given a circle with centre at the origin, which passes through the point ( 2,-1). Find its equation.

(4)Find the unit vector in the direction of vector b = 3i + 4j -5k

(5)A line AB passes through the point P (3, -2) with gradient дљН 2. Determine the equation of the line CD through P perpendicular to AB

(6)Determine the equation of a straight line passing through the point (1, 0) and (2, -3).

(7)Determine the scalar product of vectors 2i+3j−5k2i+3j−5k and 4i+j−6k

(8)Find the equation of the line which is parallel to the line 2y + 3x = 3 and passes through the midpoint of (-2,3) and (4, 5).

(9)Find the centre and radius of each of the circle x2+y2 дљН2xдљН6y=15

(10)Determine the direction cosines [l, m, n] of the vector 3i -2 j +6k

Answer 1

Answer on Question #83337 – Math – Analytic Geometry

Question

(1) The scalar product of vectors $\mathbf{a}$ and $\mathbf{b}$, where $\theta$ is the angle between them, is

Solution

$$a \cdot b = \|a\| \|b\| \cos \theta$$

Question

(2) Determine the gradient of a straight line passing through the point $(1,6)$ and $(-3,-3)$.

Solution

Gradient (slope) of a straight line

$$grad = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 6}{-3 - 1} = \frac{9}{4}$$

Question

(3) Given a circle with centre at the origin, which passes through the point $(2,-1)$. Find its equation.

Solution

The equation of the circle with centre at the origin

$$x^2 + y^2 = R^2$$

Substitute

$$(2)^2 + (-1)^2 = R^2$$ $$R^2 = 5$$

The equation of the circle with centre at the origin, which passes through the point $(2,-1)$ is

$$x^2 + y^2 = 5$$

Question

(4) Find the unit vector in the direction of vector $\mathbf{b} = 3\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}$

Solution

$$\|\mathbf{b}\| = \sqrt{(3)^2 + (4)^2 + (-5)^2} = 5\sqrt{2}$$

The unit vector in the direction of vector $\mathbf{b} = 3\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}$

$$\boldsymbol {u} = \frac {\boldsymbol {b}}{\| \boldsymbol {b} \|}$$ $$\boldsymbol {u} = \frac {3 \sqrt {2}}{1 0} \boldsymbol {i} + \frac {2 \sqrt {2}}{5} \boldsymbol {j} - \frac {\sqrt {2}}{2} \boldsymbol {k}$$

Question

(5) A line $AB$ passes through the point $P(3, -2)$ with gradient $-2$. Determine the equation of the line $CD$ through $P$ perpendicular to $AB$.

Solution

If two lines are perpendicular, then $grad_{1}grad_{2} = -1$.

$$grad_{CD} = - \frac {1}{grad_{AB}} = - \frac {1}{- 2} = \frac {1}{2}$$

The equation of the line $CD$

$$y = \frac {1}{2} x + b$$

Substitute and find $b$

$$- 2 = \frac {1}{2} (3) + b \Rightarrow b = - \frac {7}{2}$$

The equation of the line $CD$

$$y = \frac {1}{2} x - \frac {7}{2}$$

Question

(6) Determine the equation of a straight line passing through the point $(1,0)$ and $(2, -3)$.

Solution

Gradient (slope) of a straight line

$$grad = \frac {y _ {2} - y _ {1}}{x _ {2} - x _ {1}} = \frac {- 3 - 0}{2 - 1} = - 3$$

The equation of the line

$$y = - 3 x + b$$

Substitute and find $b$

$$0 = - 3 (1) + b \Rightarrow b = 3$$

The equation of a straight line passing through the point $(1,0)$ and $(2, -3)$.

$$y = - 3 x + 3$$

Question

(7) Determine the scalar product of vectors $2\boldsymbol{i} + 3\boldsymbol{j} - 5\boldsymbol{k}$ and $4\boldsymbol{i} + \boldsymbol{j} - 6\boldsymbol{k}$

**Solution**

$$(2\boldsymbol{i} + 3\boldsymbol{j} - 5\boldsymbol{k}) \cdot (4\boldsymbol{i} + \boldsymbol{j} - 6\boldsymbol{k}) = 2(4) + 3(1) + (-5)(-6) = 41$$

**Question**

(8) Find the equation of the line which is parallel to the line $2y + 3x = 3$ and passes through the midpoint of $(-2,3)$ and $(4,5)$.

**Solution**

If two lines are parallel, then they have the same gradient (slope).

Find the gradient slope

$$\begin{array}{l} 2y + 3x = 3 \Rightarrow y = -\frac{3}{2}x + \frac{3}{2} \\ grad = -\frac{3}{2} \end{array}$$

The midpoint of two points

$$x_m = \frac{x_1 + x_2}{2}, \quad y_m = \frac{y_1 + y_2}{2}$$

Substitute

$$x_m = \frac{-2 + 3}{2} = \frac{1}{2}, \quad y_m = \frac{3 + 5}{2} = 4$$

Find the equation of the line

$$y = -\frac{3}{2}x + b$$

Substitute and find $b$

$$4 = -3\left(\frac{1}{2}\right) + b \Rightarrow b = \frac{15}{2}$$

The equation of the line which is parallel to the line $2y + 3x = 3$ and passes through the midpoint of $(-2,3)$ and $(4,5)$

$$y = -\frac{3}{2} + \frac{15}{2}$$

**Question**

(9) Find the centre and radius of each of the circle $x^2 + y^2 - 2x - 6y = 15$

**Solution**

$$\begin{array}{l} x^2 + y^2 - 2x - 6y = 15 \\ x^2 - 2x + 1 + y^2 - 6y + 9 - 1 - 9 = 15 \\ (x - 1)^2 + (y - 3)^2 = 25 \end{array}$$

The equation of the circle with center at $(h,k)$ and radius $r$

$$(x - h)^2 + (y - k)^2 = r^2$$

Therefore, we have the equation of the circle with the center $(1,3)$ and radius 5.

Question

(10) Determine the direction cosines $[l, m, n]$ of the vector $3i - 2j + 6k$

Solution

Let $\mathbf{a} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}$

$$\| \mathbf {a} \| = \sqrt {(3) ^ {2} + (- 2) ^ {2} + (6) ^ {2}} = 7$$

Determine the direction cosines

$$l = \cos \alpha = \frac {x}{\| \boldsymbol {a} \|} = \frac {1}{7}$$ $$m = \cos \beta = \frac {y}{\| \boldsymbol {a} \|} = - \frac {2}{7}$$ $$n = \cos \gamma = \frac {z}{\| \boldsymbol {a} \|} = \frac {6}{7}$$

Check

$$l ^ {2} + m ^ {2} + n ^ {2} = \left(\frac {1}{7}\right) ^ {2} + \left(- \frac {2}{7}\right) ^ {2} + \left(\frac {6}{7}\right) ^ {2} = 1$$ $$[ l, m, n ] = \left[ \begin{array}{c} 1 \\ 7 \end{array} , - \begin{array}{c} 2 \\ 7 \end{array} , \begin{array}{c} 6 \\ 7 \end{array} \right].$$

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