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With respect to the origin $O$, the points $A$ and $B$ have position vectors given by $\overrightarrow {OA}  = i + 2j + 2k$ and $\overrightarrow {OB}  = 3i + 4j$. The point $P$ lies on the line $AB$ and $OP$ is perpendicular to $AB$.

a) Find a vector equation for the line $AB$.

b) Find the position vector of $P$.

c) Find the equation of the plane which contains $AB$ and which is perpendicular to the plane $OAB$, giving your answer in the form $\alpha x + by + cz = d$.

پاسخ تشریحی :
نمایش پاسخ

a) State correct equation in any form, e.g. $r = i + 2j + 2k + \lambda \left( {2i + 2j - 2k} \right)$

b) EITHER: Equate a relevant scalar product to zero and form an equation in $\lambda $

OR 1: Equate derivative of $O{P^2}$ (or $OP$) to zero and form an equation in $\lambda $

OR 2: Use Pythagoras in $OAP$ or $OBP$ and form an equation in $\lambda $

State a correct equation in any form

Solve and obtain $\lambda  =  - \frac{1}{6}$ or equivalent

Obtain final answer $\overrightarrow {OP}  = \frac{2}{3}i + \frac{5}{3}j + \frac{7}{3}k$, or equivalent

c) EITHER: State or imply $\overrightarrow {OP} $ is a normal to the required plane

State normal vector $2i + 5j + 7k$, or equivalent

Substitute coordinates of a relevant point in $2x + 5y + 7z = d$ and evaluate $d$

Obtain answer $2x + 5y + 7z = 26$, or equivalent

OR 1: Find a vector normal to plane $AOB$ and calculate its vector product with a direction vector for the line $AB$

Obtain answer $2i + 5j + 7k$, or equivalent

Substitute coordinates of a relevant point in $2x + 5y + 7z = d$ and evaluate $d$

Obtain answer $2x + 5y + 7z = 26$, or equivalent

OR 2: Set up and solve simultaneous equations in $\alpha $, $b$, c derived from zero scalar products of $\alpha i + bj + ck$ with (a) a direction vector for line $AB$, (b) a normal to plane $OAB$

Obtain $\alpha {\text{ }}:{\text{ }}b{\text{ }}:{\text{ }}c = 2{\text{ }}:{\text{ }}5{\text{ }}:{\text{ }}7$, or equivalent

Substitute coordinates of a relevant point in $2x + 5y + 7z = d$ and evaluate $d$

Obtain answer $2x + 5y + 7z = 26$, or equivalent

OR 3: With $Q{\text{ }}\left( {x,{\text{ }}y,{\text{ }}z} \right)$ on plane, use Pythagoras in $OPQ$ to form an equation in $x$, $y$ and $z$

Form a correct equation

Reduce to linear form

Obtain answer $2x + 5y + 7z = 26$, or equivalent

OR 4: Find a vector normal to plane $AOB$ and form a 2-parameter equation with relevant vectors, e.g., $r = i + 2j + 2k + \lambda \left( {2i - 2j + 2k} \right) + \mu \left( {8i - 6j + 2k} \right)$

State three correct equations in $x$, $y$, $z$, $\lambda $ and $\mu $

Eliminate $\lambda $ and $\mu $

Obtain answer $2x + 5y + 7z = 26$, or equivalent

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