The line $l_1$ has vector equation $$ extbf{r} = 8 extbf{i} + 12 extbf{j} + 14 extbf{k} + \lambda ( extbf{i} - extbf{j} - extbf{k}) $$ where $\lambda$ is a parameter. The point A has coordinates (4, 8, $a$), where $a$ is a constant. The point B has coordinates ($b$, 13, 13), where $b$ is a constant. Points A and B lie on the line $l_1$. (a) Find the values of $a$ and $b$. Given that the point O is the origin, and that the point P lies on $l_1$ such that OP is perpendicular to $l_1$, (b) find the coordinates of P. (c) Hence find the distance OP, giving your answer as a simplified surd.

Photo AI

The line $l_1$ has vector equation $$ extbf{r} = 8 extbf{i} + 12 extbf{j} + 14 extbf{k} + \lambda ( extbf{i} - extbf{j} - extbf{k}) $$ where $\lambda$ is a parameter - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 7

Question 8

$The-line-$l_1$-has-vector-equation--$$--extbf{r}-=-8-extbf{i}-+-12-extbf{j}-+-14-extbf{k}-+-\lambda-(--extbf{i}----extbf{j}----extbf{k})-$$--where-$\lambda$-is-a-parameter-Edexcel-A-Level Maths Pure-Question 8-2006-Paper 7.png$

The line $l_1$ has vector equation $$ extbf{r} = 8 extbf{i} + 12 extbf{j} + 14 extbf{k} + \lambda ( extbf{i} - extbf{j} - extbf{k}) $$ where $\lambda$ is a par... show full transcript

Worked Solution & Example Answer:The line $l_1$ has vector equation $$ extbf{r} = 8 extbf{i} + 12 extbf{j} + 14 extbf{k} + \lambda ( extbf{i} - extbf{j} - extbf{k}) $$ where $\lambda$ is a parameter - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 7

Step 1

(a) Find the values of $a$ and $b$

96%

114 rated

Answer

To find the values of $a$ and $b$ , we first substitute the coordinates of point A and point B into the line equation.

For point A, we have: $(4, 8, a) = (8, 12, 14) + \lambda (1, -1, -1)$

This gives us the system of equations:

$8 + \lambda = 4$
$12 - \lambda = 8$
$14 - \lambda = a$

Solving the first equation, we find: $\lambda = 4 - 8 = -4$

Now substituting $\lambda = -4$ into the second equation, we find: $12 - (-4) = 8$

This is valid, and now substituting $\lambda = -4$ into the third equation: $14 - (-4) = a \implies a = 14 + 4 = 18$

Now for point B, substituting coordinates ( $b$ , 13, 13): $(b, 13, 13) = (8, 12, 14) + \mu (1, -1, -1)$

Giving us:

$8 + \mu = b$
$12 - \mu = 13$
$14 - \mu = 13$

From the second equation: $\mu = 12 - 13 = -1$

Now substituting $\mu = -1$ into the first equation: $8 - 1 = b \implies b = 8 - 1 = 7$

Thus, we have found:

$a = 18$
$b = 7$ .

Step 2

(b) find the coordinates of P

99%

104 rated

Answer

Since OP is perpendicular to the line $l_1$ , we need to use the direction ratios of the line to establish the relationship. The direction ratios for line $l_1$ are from the vector $(1, -1, -1)$ : $\textbf{d} = (1, -1, -1)$

Let the coordinates of point P be $(x_p, y_p, z_p)$ , and since P lies on the line we can express it as: $(x_p, y_p, z_p) = (8, 12, 14) + \lambda (1, -1, -1)$

Thus:

$x_p = 8 + \lambda$
$y_p = 12 - \lambda$
$z_p = 14 - \lambda$

Since OP is perpendicular to $l_1$ , we have: $\textbf{OP} \cdot \textbf{d} = 0$

The vector OP is given by: $\textbf{OP} = (x_p, y_p, z_p) - (0, 0, 0) = (x_p, y_p, z_p)$

Now substituting from our equations: $(8 + \lambda, 12 - \lambda, 14 - \lambda) \cdot (1, -1, -1) = 0$

This leads to: $(8 + \lambda) - (12 - \lambda) - (14 - \lambda) = 0 \implies \lambda + 8 - 12 - 14 = 0$

Solving yields: $\lambda = 18$

Now substituting $\lambda$ back into the equations for P gives:

$x_p = 8 + 18 = 26$
$y_p = 12 - 18 = -6$
$z_p = 14 - 18 = -4$

Thus, the coordinates of P are: $P(26, -6, -4)$ .

Step 3

(c) Hence find the distance OP, giving your answer as a simplified surd

96%

101 rated

Answer

To find the distance OP, we can use the distance formula:

$OP = \sqrt{(x_p - 0)^2 + (y_p - 0)^2 + (z_p - 0)^2}$

Substituting the coordinates of P: $OP = \sqrt{(26 - 0)^2 + (-6 - 0)^2 + (-4 - 0)^2}$ $= \sqrt{26^2 + (-6)^2 + (-4)^2}$ $= \sqrt{676 + 36 + 16}$ $= \sqrt{728}$

We can factor this as: $= \sqrt{36 \cdot 20.222...} = 6\sqrt{20.222...} = 14\sqrt{2}$

Hence the distance OP is: $OP = 14\sqrt{2}$ .

The line $l_1$ has vector equation $$ extbf{r} = 8 extbf{i} + 12 extbf{j} + 14 extbf{k} + \lambda ( extbf{i} - extbf{j} - extbf{k}) $$ where $\lambda$ is a parameter - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 7

Worked Solution & Example Answer:The line $l_1$ has vector equation $$ extbf{r} = 8 extbf{i} + 12 extbf{j} + 14 extbf{k} + \lambda ( extbf{i} - extbf{j} - extbf{k}) $$ where $\lambda$ is a parameter - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 7

(a) Find the values of $a$ and $b$

(b) find the coordinates of P

(c) Hence find the distance OP, giving your answer as a simplified surd

Join the A-Level students using SimpleStudy...