Help with Geometry of Vectors Questions

[Guest2]

View attachment 5379

I would appreciate any help with this questions because I truly horrid at geometry questions.

I've only done (i) to which I've found the answer to be (E). I can't do from from part (ii) on.

 

[Opalg]

A couple of hints to get you started:

For (ii), this plane is the perpendicular bisector of the line $AB$.

For (iii), use the cross product $\mathbf{a} \times \mathbf{b}$. This has magnitude $|\mathbf{a}|\,|\mathbf{b}|\sin\theta$, where $\theta$ is the angle between $OA$ and $OB$.

 

[Guest2]

A couple of hints to get you started:

For (ii), this plane is the perpendicular bisector of the line $AB$.

For (iii), use the cross product $\mathbf{a} \times \mathbf{b}$. This has magnitude $|\mathbf{a}|\,|\mathbf{b}|\sin\theta$, where $\theta$ is the angle between $OA$ and $OB$.

Thank you.

The midpoint between the two points will be a point on the plane. This point is $\left(\frac{4-2}{2}, \frac{1+7}{2}, \frac{4-2}{2} \right) = (1,4,1)$.

The direction vector between the two points, which is also a normal to the plane, is $(-2-4, 7-1, 4+6) = (-6, 6, 6)$

The general equation for a plane is $\mathbf{n} \cdot <x-x_0, y-y_0, z-z_0> =0 $, we have $\mathbf{n} = (-6,6,6)$ and $(x_0, y_0, z_0) =(1,4,1)$

So we have $<1,4,1><-6, 6, 6> = 0 \implies x-y-z = -4$ and so the answer is (B).

I just have a question that's due to my ignorance. I calculated the direction vector as the difference $\mathbf{b}-\mathbf{a}$. But how do we know which to subtract from which vector when finding the direction vector? I imagine I would have got a different answer if I calculated the direction vector as $\mathbf{a}-\mathbf{b}$.

 

[Guest2]

For (iii) we have $\mathbf{a} \times \mathbf{b} = (18, -12, 30) \implies ||\mathbf{a} \times \mathbf{b}|| = \sqrt{1368} \implies \frac{1}{2}||\mathbf{a} \times \mathbf{b}|| = \sqrt{342} \implies \triangle ^2 = 342$.

The reason being the area of the triangle is half the area of the parallelogram formed which is $||\mathbf{a} \times \mathbf{b}||$. (Happy)

So the answer to $(iii)$ is $(A)$. If anyone could give me a hint or suggestion for part $(iv)$ I'd appreciate it.

 

[Guest2]

For $(v)$ we have $ (\mathbf{a}-\mathbf{b}) \cdot \mathbf{w}(t) = 0 \implies <6,-6,-6><2, t, 1> = 0\implies 6t = 6 \implies t = 1$, so the answer to (v) is $(C)$.

For $(iv)$ let $P = (a, b, c)$ now $|PA| = 2|PB| \iff [(4,1,2)-(a, b, c)] = 2[(-2,7,4) -(a, b, c)] \iff (a, b, c) = (-8, 13, 10).$

So is the answer to $(iv) ~(F)$ none of these, or am I getting something wrong (more likely)? Only this remains now! (Happy)

 

[Opalg]

For $(v)$ we have $ (\mathbf{a}-\mathbf{b}) \cdot \mathbf{w}(t) = 0 \implies <6,-6,-6><2, t, 1> = 0\implies 6t = 6 \implies t = 1$, so the answer to (v) is $(C)$.

For $(iv)$ let $P = (a, b, c)$ now $|PA| = 2|PB| \iff [(4,1,2)-(a, b, c)] = 2[(-2,7,4) -(a, b, c)] \iff (a, b, c) = (-8, 13, 10).$

So is the answer to $(iv) ~(F)$ none of these, or am I getting something wrong (more likely)? Only this remains now! (Happy)

For (iv), $P$ is two-thirds of the way from $A$ to $B$. So its position vector will be $\frac23\mathbf{b} + \frac13\mathbf{a}.$ Your answer $(-8,13,10)$ satisfies the condition $|PA| = 2|PB|$, but it does not lie on the line segment $AB$. In fact is lies beyond $B$, in the opposite direction from $A$.

 

[suluclac]

Part (i) should be easy; 4(-2) + 1(7) + -2(4) = -9.