We read your photo as:

In the diagram below, OAB is a triangle in which OA = 3g and OB = 6b. OC : CA = 2 : 3 and AD : DB = 1 : 2. OD meets CB at E. Express each of the following in terms of a and/or b: (a) AB, (b) OD, (c) BC. Given that BE = hBC, express BE in terms of h, a and b.

Mathematics

ScanSolve AI Answer

Step-by-step solution

10 steps

(a): Flowchart for Pseudocode

As an AI, I cannot draw a flowchart directly. However, I can describe the elements and their connections to represent the given pseudocode.

  • Start (Oval)
  • Input radius (Parallelogram)
  • Decision: radius < 0? (Diamond)
    • If Yes:
      • Display "error message" (Parallelogram)
      • Input positive radius (Parallelogram)
      • (Loop back to the "Decision: radius < 0?" diamond)
    • If No:
      • Input height (Parallelogram)
      • Decision: height < 0? (Diamond)
        • If Yes:
          • Display "error message" (Parallelogram)
          • Input positive height (Parallelogram)
          • (Loop back to the "Decision: height < 0?" diamond)
        • If No:
          • Process: Volume = (1/3) * π\pi * radius2^2 * height (Rectangle)
  • Display volume (Parallelogram)
  • Stop (Oval)

(b): Vector Geometry

Given: OA⃗=3a⃗\vec{OA} = 3\vec{a}, OB⃗=6b⃗\vec{OB} = 6\vec{b}, OC : CA = 2 : 3, AD : DB = 1 : 2.

b) (i) Express each of the following in terms of a⃗\vec{a} and/or b⃗\vec{b}

a) AB⃗\vec{AB}

  1. Use the triangle law for vector addition.

    AB⃗=AO⃗+OB⃗\vec{AB} = \vec{AO} + \vec{OB}

  2. Substitute the given vectors.

    Since AO⃗=−OA⃗\vec{AO} = -\vec{OA}, we have:

    AB⃗=−OA⃗+OB⃗=−3a⃗+6b⃗\vec{AB} = -\vec{OA} + \vec{OB} = -3\vec{a} + 6\vec{b} AB⃗=6b⃗−3a⃗\vec{AB} = \boxed{6\vec{b} - 3\vec{a}}

    b) OD⃗\vec{OD}

  3. Use the section formula for position vectors.

    Point D divides the line segment AB in the ratio AD : DB = 1 : 2.

    OD⃗=2OA⃗+1OB⃗1+2\vec{OD} = \frac{2\vec{OA} + 1\vec{OB}}{1+2}

  4. Substitute the given vectors.

    OD⃗=2(3a⃗)+1(6b⃗)3\vec{OD} = \frac{2(3\vec{a}) + 1(6\vec{b})}{3} OD⃗=6a⃗+6b⃗3\vec{OD} = \frac{6\vec{a} + 6\vec{b}}{3} OD⃗=2a⃗+2b⃗\vec{OD} = \boxed{2\vec{a} + 2\vec{b}}

    c) BC⃗\vec{BC}

  5. Find OC⃗\vec{OC} using the section formula.

    Point C divides the line segment OA in the ratio OC : CA = 2 : 3.

    OC⃗=22+3OA⃗=25OA⃗\vec{OC} = \frac{2}{2+3}\vec{OA} = \frac{2}{5}\vec{OA}

  6. Substitute OA⃗\vec{OA}.

    OC⃗=25(3a⃗)=65a⃗\vec{OC} = \frac{2}{5}(3\vec{a}) = \frac{6}{5}\vec{a}

  7. Use the triangle law for vector addition to find BC⃗\vec{BC}.

    BC⃗=BO⃗+OC⃗\vec{BC} = \vec{BO} + \vec{OC}

  8. Substitute the vectors.

    Since BO⃗=−OB⃗\vec{BO} = -\vec{OB}, we have:

    BC⃗=−OB⃗+OC⃗=−6b⃗+65a⃗\vec{BC} = -\vec{OB} + \vec{OC} = -6\vec{b} + \frac{6}{5}\vec{a} BC⃗=65a⃗−6b⃗\vec{BC} = \boxed{\frac{6}{5}\vec{a} - 6\vec{b}}

    b) (ii) Given that BE⃗=hBC⃗\vec{BE} = h\vec{BC}, express BE⃗\vec{BE} in terms of hh, a⃗\vec{a} and b⃗\vec{b}.

  9. Substitute the expression for BC⃗\vec{BC} from part (i)(c).

    BE⃗=h(65a⃗−6b⃗)\vec{BE} = h \left( \frac{6}{5}\vec{a} - 6\vec{b} \right)

  10. Distribute hh into the expression.

    BE⃗=6h5a⃗−6hb⃗\vec{BE} = \boxed{\frac{6h}{5}\vec{a} - 6h\vec{b}}
Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp