The position vectors of points A, B and C are such that OA = 3i - 5j - 4k, OB = j + 8k and OC = -2i + 5j + 16k. Show that the points A, B and C are collinear.
|Mathematics
The position vectors of points A, B and C are such that OA = 3i - 5j - 4k, OB = j + 8k and OC = -2i + 5j + 16k. Show that the points A, B and C are collinear.
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Answer
Points A, B, and C are collinear.
Fresh day Oliverkoome@gmail254.Com., let's solve.
14. The position vectors of points A, B and C are such that OA=3i−5j−4k, OB=j+8k and OC=−2i+5j+16k. Show that the points A, B and C are collinear.
Step 1: Write the position vectors in column form.
OA=3−5−4OB=018OC=−2516
Step 2: Calculate the vector AB.
AB=OB−OA=018−3−5−4=0−31−(−5)8−(−4)=−3612
Step 3: Calculate the vector BC.
BC=OC−OB=−2516−018=−2−05−116−8=−248
Step 4: Check if AB is a scalar multiple of BC.
We need to find a scalar k such that AB=kBC.
−3612=k−248
Comparing the components:
For the x-component: −3=k(−2)⟹k=−2−3=23
For the y-component: 6=k(4)⟹k=46=23
For the z-component: 12=k(8)⟹k=812=23
Since k=23 is consistent for all components, AB=23BC.
This shows that vectors AB and BC are parallel. Since they share a common point B, the points A, B, and C are collinear.
PointsA,B,andCarecollinear.
15. Given that sin75∘=46+2, simplify sin75∘1.
Step 1: Substitute the given value of sin75∘ into the expression.
sin75∘1=46+21
Step 2: Invert the fraction in the denominator.
sin75∘1=6+24
Step 3: Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is 6−2.
6+24×6−26−2
Step 4: Perform the multiplication. Use the difference of squares formula (a+b)(a−b)=a2−b2 in the denominator.
=(6)2−(2)24(6−2)=6−24(6−2)=44(6−2)
Step 5: Simplify the expression.
=6−2
The simplified expression is:
6−2
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Fresh day Oliverkoome@gmail254.Com., let's solve. 14.
The position vectors of points A, B and C are such that OA = 3i - 5j - 4k, OB = j + 8k and OC = -2i + 5j + 16k. Show that the points A, B and C are collinear.
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Fresh day Oliverkoome@gmail254.Com., let's solve. 14. The position vectors of points A, B and C are such that OA = 3i - 5j - 4k, OB = j + 8k and OC = -2i + 5j + 16k. Show that the points A, B and C are collinear. Step 1: Write the position vectors in column form. OA = 3 \\ -5 \\ -4 OB = 0 \\ 1 \\ 8 OC = -2 \\ 5 \\ 16 Step 2: Calculate the vector AB. AB = OB - OA = 0 \\ 1 \\ 8 - 3 \\ -5 \\ -4 = 0 - 3 \\ 1 - (-5) \\ 8 - (-4) = -3 \\ 6 \\ 12 Step 3: Calculate the vector BC. BC = OC - OB = -2 \\ 5 \\ 16 - 0 \\ 1 \\ 8 = -2 - 0 \\ 5 - 1 \\ 16 - 8 = -2 \\ 4 \\ 8 Step 4: Check if AB is a scalar multiple of BC. We need to find a scalar k such that AB = k BC. -3 \\ 6 \\ 12 = k -2 \\ 4 \\ 8 Comparing the components: For the x-component: -3 = k(-2) k = (-3)/(-2) = (3)/(2) For the y-component: 6 = k(4) k = (6)/(4) = (3)/(2) For the z-component: 12 = k(8) k = (12)/(8) = (3)/(2) Since k = (3)/(2) is consistent for all components, AB = (3)/(2) BC. This shows that vectors AB and BC are parallel. Since they share a common point B, the points A, B, and C are collinear. Points A, B, and C are collinear. 15. Given that 75^ = sqrt(6) + sqrt(2)4, simplify (1)/( 75^). Step 1: Substitute the given value of 75^ into the expression. (1)/( 75^) = (1)/(sqrt(6) + 2)4 Step 2: Invert the fraction in the denominator. (1)/( 75^) = (4)/(sqrt(6) + 2) Step 3: Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is sqrt(6) - sqrt(2). (4)/(sqrt(6) + 2) × sqrt(6) - sqrt(2)sqrt(6) - sqrt(2) Step 4: Perform the multiplication. Use the difference of squares formula (a+b)(a-b) = a^2 - b^2 in the denominator. = 4(sqrt(6) - sqrt(2))(sqrt(6))^2 - (sqrt(2))^2 = 4(sqrt(6) - sqrt(2))6 - 2 = 4(sqrt(6) - sqrt(2))4 Step 5: Simplify the expression. = sqrt(6) - sqrt(2) The simplified expression is: sqrt(6) - sqrt(2) What's next?