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The line L1 has vector equation,  L1 = (  6, 1 ,-1  ) + λ ( 2, 1, 0). The line L2 passes through the points (2, 3, −1) and (4, −1, 1). i) find vector equation of L2 ii)show L2 and L1 are perpendicular.

i)L(r) vector line equations in general are in the form L(r) = p1 + λ (p2) where p1 is any ponit on the line and p2 is the vector direction of the line(unsure how to get the whiteboard up or i would describe...
CE
Answered by Charles E. • Maths tutor
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The straight line with equation y=3x-7 does not cross or touch the curve with equation y=2px^2-6px+4p, where p is a constant.(a) Show that 4p^2-20p+9<0 (b) Hence find the set of possible values for p.

(a) If we consider the intersection of these two lines then, 3x-7=2px^2-6px+4p.This can be rearranged into the form ax^2+bx+c=0 such that, 2px^2+(-6p-3)x+4p+7, where a=2p, b=-6p-3 and c=4p+7.However, since t...
WG
Answered by Will G. • Maths tutor
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How do you differentiate using the chain rule?

The chain rule is used where the equation you are looking to differentiate is a function that is itself raised to a power. For example, we might have y = (x 2 -2) 3 and want to differentiate with respect to ...
OB
Answered by Oliver B. • Maths tutor
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C and D are two events such that P(C) = 0.2, P(D) = 0.6 and P(C|D) = 0.3. Find P(D|C), P(C’ ∩ D’) & P(C’ ∩ D)

Here we use the formula for conditional probability:P(C|D)xP(D)=(𝑃(C∩D))= 0.3x0.6=0.18Note: 𝑃(C∩D)=𝑃(D∩C)=0.18Hence P(D|C)=𝑃(D∩C)/P(C) = 0.18/0.2 = 0.9P(C'∩D') = P(C'|D')xP(D')P(C'|D) = 0.7P(D')=0.4P(C')=0.8...
SV
Answered by Sunny V. • Maths tutor
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Given that x = cot y, show that dy/dx = -1/(1+x^2)

Identify that we are looking at dy/dx, not dx/dy and realise the relationship that dy/dx=1/(dx/dy)2)Try find dx/dy;cot = 1/tan or (tan) -1 Hence, x=(tan y ) -1 implying dx/dy = (-1)(tan y) -2 (sec 2 y ) =(-1...
JF
Answered by Jacob F. • Maths tutor
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