Question

(1) If r = 2i дљН 4j, s = 2i + 6j, t = 3i дус j, find 2r-t+s

(2)Find the equation of the tangent to the curve y=x3−x2y=x3−x2 at the point (1, 0).

(3)Find the area of a triangle with vertices (3, 1), (0, 0) and (1, 2).

(4)If the point C(2,−1) be the centre of a circle that passes through the point A(-2,2) find the equation of the cycle

(5)If r = 2i дљН 4j, s = 2i + 6j, t = 3i дус j, find magnitude of the vector 2r-t+s

(6)P, Q and R are points (1, -6), (3, 6) and (5, 2) respectively. Determine the length of the line joining the mid point of PQ and QR

(7)Find the equation of a straight line passing through the points (2, 3) and (-2, 5).

(8)Let Z1=6iдљН4j+4kZ1=6iдљН4j+4k, Z2=i+6j−kZ2=i+6j−k, find magnitude of th

(9)Find the equation of the normal to the curve y=x3−x2y=x3−x2 at the point (1, 1

Answer 1

Answer on Question #83335 – Math – Analytic Geometry

Question

(1) If $r = 2i - 4j$, $s = 2i + 6j$, $t = 3i - j$, find $2r - t + s$

Solution

$$2r - t + s = 2(2i - 4j) - (3i - j) + 2i + 6j = 3i - j$$

Question

(2) Find the equation of the tangent to the curve $y = x^3 - x^2$ at the point $(1,0)$.

Solution

Find the first derivative with respect to $x$

$$y' = (x^3 - x^2)' = 3x^2 - 2x$$

Find the slope of the tangent line

$$\text{slope} = m = y'(1) = 3(1)^2 - 2(1) = 1$$

The equation of the tangent to the curve in point slope form

$$y - 0 = 1(x - 1)$$

The equation of the tangent to the curve in slope-intercept form

$$y = x - 1$$

Question

(3) Find the area of a triangle with vertices $(3,1)$, $(0,0)$ and $(1,2)$.

Solution

Let $A(3,1), O(0,0)$, and $B(1,2)$ be the vertices of a triangle.

$$\overrightarrow{OA} = (x_A - x_O, y_A - y_O) = (3 - 0, 1 - 0) = (3, 1)$$ $$\overrightarrow{OB} = (x_B - x_O, y_B - y_O) = (1 - 0, 2 - 0) = (1, 2)$$

Find the cross product

$$\overrightarrow{OA} \times \overrightarrow{OB} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 3 & 1 & 0 \\ 1 & 2 & 0 \end{vmatrix} = \vec{i} \begin{vmatrix} 1 & 0 \\ 2 & 0 \end{vmatrix} - \vec{j} \begin{vmatrix} 3 & 0 \\ 1 & 0 \end{vmatrix} + \vec{k} \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} = 0 \vec{i} - 0 \vec{j} + 5 \vec{k}$$ $$S_{\Delta ABO} = \frac{1}{2} \left\| \overrightarrow{OA} \times \overrightarrow{OB} \right\| = \frac{1}{2}(5) = \frac{5}{2} \text{ (units}^2\text{)}$$

Question

(4) If the point $C(2, -1)$ be the centre of a circle that passes through the point $A(-2, 2)$ find the equation of the circle.

Solution

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

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

Substitute and find $r$

$$\begin{array}{l} (-2 - 2)^2 + (2 - (-1))^2 = r^2 \\ r^2 = 25 \Rightarrow r = 5 \\ \end{array}$$

The equation of the circle with center at $C(2, -1)$ and radius 5

$$(x - 2)^2 + (y + 1)^2 = 25$$

Question

(5) If $r = 2i - 4j$, $s = 2i + 6j$, $t = 3i - j$, find magnitude of the vector $2r - t + s$

Solution

$2r - t + s = 2(2i - 4j) - (3i - j) + 2i + 6j = 3i - j$

Find magnitude of the vector $2r - t + s$

$$\|2r - t + s\| = \sqrt{(3)^2 + (-1)^2} = \sqrt{10}$$

Question

(6) $P, Q$ and $R$ are points $(1, -6), (3, 6)$ and $(5, 2)$ respectively. Determine the length of the line joining the midpoint of $PQ$ and $QR$.

Solution

The midpoint of two points

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

The midpoint of $PQ$

$$x_{mPQ} = \frac{1 + 3}{2} = 2, \quad y_{mPQ} = \frac{-6 + 6}{2} = 0$$

The midpoint of $QR$

$$x_{mQR} = \frac{3 + 5}{2} = 4, \quad y_{mQR} = \frac{6 + 2}{2} = 4$$

Determine the length of the line joining the midpoint of $PQ$ and $QR$.

$$l = \sqrt{(4 - 2)^2 + (4 - 0)^2} = 2\sqrt{5}$$

Question

(7) Find the equation of a straight line passing through the points $(2, 3)$ and $(-2, 5)$.

Solution

Gradient (slope) of a straight line

$$grad = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{-2 - 2} = -\frac{1}{2}$$

The equation of the line

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

Substitute and find $b$

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

The equation of a straight line passing through the points (2,3) and $(-2,5)$.

$$y = -\frac{1}{2}x + 4$$

Question

(8) Let $Z_1 = 6\vec{i} - 4\vec{j} + 4\vec{k}$, $Z_2 = \vec{i} + 6\vec{j} - \vec{k}$, find magnitude of the cross product $\overrightarrow{Z_1} \times \overrightarrow{Z_2}$.

Solution

Find the cross product $\overrightarrow{Z_1} \times \overrightarrow{Z_2}$

$$\begin{array}{l} \overrightarrow{Z_1} \times \overrightarrow{Z_2} = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 6 & -4 & 4 \\ 1 & 6 & -1 \end{array} \right| = \vec{i} \left| \begin{array}{cc} -4 & 4 \\ 6 & -1 \end{array} \right| - \vec{j} \left| \begin{array}{cc} 6 & 4 \\ 1 & -1 \end{array} \right| + \vec{k} \left| \begin{array}{cc} 6 & -4 \\ 1 & 6 \end{array} \right| = \\ = \vec{i} \left(-4(-1) - 4(6)\right) - \vec{j} \left(6(-1) - 4(1)\right) + \vec{k} \left(6(6) - (-4)(1)\right) = \\ = -20\vec{i} + 10\vec{j} + 40\vec{k} \end{array}$$

Find magnitude of the cross product $\overrightarrow{Z_1} \times \overrightarrow{Z_2}$

$$\left\| \overrightarrow{Z_1} \times \overrightarrow{Z_2} \right\| = \sqrt{(-20)^2 + (10)^2 + (40)^2} = 10\sqrt{21}$$

Question

(9) Find the equation of the normal to the curve $y = x^3 - x^2$ at the point $(1,1)$.

Solution

Find the first derivative with respect to $x$

$$y' = (x^3 - x^2)' = 3x^2 - 2x$$

Find the slope of the normal line

$$slope = m = -\frac{1}{y'(1)} = -\frac{1}{3(1)^2 - 2(1)} = -1$$

The equation of the normal to the curve in point slope form

$$y - 1 = -1(x - 1)$$

The equation of the normal to the curve in slope-intercept form

$$y = -x + 2$$

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