## (i) METHOD 1

State $B = (-1, -7, 2) + t(1, 2, -2)$

On plane $\Rightarrow (-1-t) + 2(-7+2t) - 2(2-2t) = -1$

$\Rightarrow t = 2 \Rightarrow B = (1, -3, -2)$

$AB = \sqrt{2^2 + 4^2 + 4^2}$ OR $2\sqrt{1^2 + 2^2 + 2^2} = 6$

| M1 | For using vector normal to plane |
| M1 | For substituting parametric form into plane |
| M1 | For solving a linear equation in $t$ |
| A1 | For correct coordinates |
| A1 | For correct length of $AB$ (5 marks) |

## (i) METHOD 2

$AB = \left|\frac{|-1-14-4+1|}{\sqrt{1^2 + 2^2 + 2^2}}\right| = 6$

OR $AB = AC \cdot \overrightarrow{AB} = \frac{|6, 7, 1| \cdot |1, 2, -2|}{\sqrt{1^2 + 2^2 + 2^2}} = 6$

$B = (-1, -7, 2) \pm 6\frac{(2, 4, -4)}{\sqrt{1^2 + 2^2 + 2^2}}$

$B = (-1, -7, 2) \pm (2, 4, -4)$

$B = (1, -3, -2)$

| M1 | For using a correct distance formula |
| A1 | For correct length of $AB$ |
| M1 | For using $B = A + $ length of $AB \times$ unit normal |
| B1 | For checking whether $+$ or $-$ is needed (substitute into plane equation) |
| A1 | For correct coordinates (allow if B0) |

## (ii)

Find vector product of any two of $4[6, 7, 1], 1[6, -3, 0], 1(0, 10, 1)$

Obtain $\mathbf{n}[1, 2, -20]$

$\theta = \cos^{-1}\frac{|[1,2,-2]| \cdot |1, 2, -20|}{\sqrt{1^2 + 2^2 + 2^2}\sqrt{1^2 + 2^2 + 20^2}}$

| M1 | For finding vector product of two relevant vectors |
| A1 | For correct vector $\mathbf{n}$ |
| M1* | For using scalar product of two normal vectors |
| M1 | For stating both moduli in denominator (dep*) |
| A1 | For correct scalar product, f.t. from $\mathbf{n}$ |

$\theta = \cos^{-1}\frac{45}{\sqrt{9}\sqrt{405}} = 41.8°$ (41.810..., 0.72972...)

| A1 | For correct angle (6 marks) |

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