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Practice: vectors in the geometric approach

Practice: vectors in the geometric approach

1

ABCDEK

ABCD\displaystyle ABCD is a parallelogram, AB→=u‾\displaystyle {\overrightarrow{AB}}={\underline{u}}, AD→=v‾\displaystyle {\overrightarrow{AD}}={\underline{v}}.
E\displaystyle E is on CD\displaystyle CD, and DE:EC=1:2\displaystyle {DE:EC}={1:2}. The segments AE\displaystyle AE and BD\displaystyle BD meet at the point K\displaystyle K.

a
AE→\displaystyle \overrightarrow{AE}

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

    b

    Express AK→\displaystyle \overrightarrow{AK} in two ways, compare coefficients and solve.

      c

      In what ratio does K\displaystyle K divide AE\displaystyle AE?

      2

      ABCDEP

      In the triangle ABC\displaystyle ABC: AB→=u‾\displaystyle {\overrightarrow{AB}}={\underline{u}}, AC→=v‾\displaystyle {\overrightarrow{AC}}={\underline{v}}.
      D\displaystyle D is the midpoint of AB\displaystyle AB. E\displaystyle E is on AC\displaystyle AC, and AE:EC=2:1\displaystyle {AE:EC}={2:1}. The segments BE\displaystyle BE and CD\displaystyle CD meet at the point P\displaystyle P.

      a
      AP→=AB→+s⋅BE→\displaystyle {\overrightarrow{AP}}={\overrightarrow{AB}+s\cdot\overrightarrow{BE}}

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

        b
        AP→=AC→+t⋅CD→\displaystyle {\overrightarrow{AP}}={\overrightarrow{AC}+t\cdot\overrightarrow{CD}}

          / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

          / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

          c

          Compare coefficients and solve.

            d

            Find the ratios BP:PE\displaystyle BP:PE and CP:PD\displaystyle CP:PD.

            3

            In the box ABCDA′B′C′D′\displaystyle ABCDA'B'C'D': AB→=u‾\displaystyle {\overrightarrow{AB}}={\underline{u}}, AD→=v‾\displaystyle {\overrightarrow{AD}}={\underline{v}}, AA′→=w‾\displaystyle {\overrightarrow{AA'}}={\underline{w}}.
            K\displaystyle K is the midpoint of A′D′\displaystyle A'D', and L\displaystyle L the centre of the face BCC′B′\displaystyle BCC'B'.

            a
            AK→\displaystyle \overrightarrow{AK}

              / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

              / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

              b
              AL→\displaystyle \overrightarrow{AL}

                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                c
                KL→\displaystyle \overrightarrow{KL}

                  / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                  / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                  d

                  Is KL\displaystyle KL parallel to the plane ABB′A′\displaystyle ABB'A'? Explain.

                  4

                  ABCDA′B′C′D′

                  In the cube ABCDA′B′C′D′\displaystyle ABCDA'B'C'D': AB→=a‾\displaystyle {\overrightarrow{AB}}={\underline{a}}, AD→=b‾\displaystyle {\overrightarrow{AD}}={\underline{b}}, AA′→=c‾\displaystyle {\overrightarrow{AA'}}={\underline{c}}.
                  The face diagonals from A\displaystyle A: v1‾=AC→\displaystyle {\underline{v_1}}={\overrightarrow{AC}}, v2‾=AB′→\displaystyle {\underline{v_2}}={\overrightarrow{AB'}}, v3‾=AD′→\displaystyle {\underline{v_3}}={\overrightarrow{AD'}}.

                  a

                  Is v3‾\displaystyle \underline{v_3} linearly dependent on v1‾\displaystyle \underline{v_1} and v2‾\displaystyle \underline{v_2}? Try v3‾=sv1‾+tv2‾\displaystyle {\underline{v_3}}={s\underline{v_1}+t\underline{v_2}}, compare the coefficients of a‾\displaystyle \underline{a}, b‾\displaystyle \underline{b}, c‾\displaystyle \underline{c}, and solve.

                    b
                    Show that 12(v1‾+v2‾+v3‾)=AC′→\displaystyle {\frac{1}{2}(\underline{v_1}+\underline{v_2}+\underline{v_3})}={\overrightarrow{AC'}}: write the left side with a‾\displaystyle \underline{a}, b‾\displaystyle \underline{b}, c‾\displaystyle \underline{c} and simplify

                      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                      5

                      OA→=u‾\displaystyle {\overrightarrow{OA}}={\underline{u}}, OB→=v‾\displaystyle {\overrightarrow{OB}}={\underline{v}}, OC→=2u‾+kv‾\displaystyle {\overrightarrow{OC}}={2\underline{u}+k\underline{v}}, where u‾\displaystyle \underline{u} and v‾\displaystyle \underline{v} are independent.
                      For which k\displaystyle k do the points A\displaystyle A, B\displaystyle B, C\displaystyle C lie on one line? Write AC→=t⋅AB→\displaystyle {\overrightarrow{AC}}={t\cdot\overrightarrow{AB}}, compare coefficients and solve.