SolveItClass 12 · NCERT

NCERT Solutions · Class 12 Mathematics Vector Algebra

73 questions · 73 still being checked

EXERCISE 10.2 1–10 (part 2 of 8)

  1. Exercise 1

    Compute the magnitude of the following vectors: a⃗=i^+j^+k;b⃗=2i^−7j^−3k^;c⃗=13i^+13j^−13k^\vec{a}=\hat{i}+\hat{j}+k ; \quad \vec{b}=2 \hat{i}-7 \hat{j}-3 \hat{k} ; \quad \vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle |\vec{a}|=\sqrt{3},|\vec{b}|=\sqrt{62},|\vec{c}|=1\)
    The magnitude of \(\displaystyle \vec{v}=x\hat{i}+y\hat{j}+z\hat{k}\) is \(\displaystyle |\vec{v}|=\sqrt{x^{2}+y^{2}+z^{2}}\).For \(\displaystyle \vec{a}=\hat{i}+\hat{j}+\hat{k}\): \[|\vec{a}|=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}\]For \(\displaystyle \vec{b}=2\hat{i}-7\hat{j}-3\hat{k}\): \[|\vec{b}|=\sqrt{2^{2}+(-7)^{2}+(-3)^{2}}=\sqrt{4+49+9}=\sqrt{62}\]For \(\displaystyle \vec{c}=\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}-\frac{1}{\sqrt{3}}\hat{k}\) the minus sign disappears on squaring: \[|\vec{c}|=\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=\sqrt{1}=1\]\(\displaystyle |\vec{a}|=\sqrt{3}\), \(\displaystyle |\vec{b}|=\sqrt{62}\), \(\displaystyle |\vec{c}|=1\) (so \(\displaystyle \vec{c}\) is a unit vector).

  2. Exercise 2

    Write two different vectors having same magnitude.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    An infinite number of possible answers.
    Two vectors are equal only when all corresponding components agree, while the magnitude \(\displaystyle \sqrt{x^{2}+y^{2}+z^{2}}\) is unchanged if the components are permuted. So permuting components produces different vectors of the same length.Take \[\vec{a}=2\hat{i}+3\hat{j}+6\hat{k},\qquad \vec{b}=6\hat{i}+3\hat{j}+2\hat{k}\] Then \[|\vec{a}|=\sqrt{4+9+36}=\sqrt{49}=7,\qquad |\vec{b}|=\sqrt{36+9+4}=\sqrt{49}=7\] They really are different vectors: the \(\displaystyle \hat{i}\)-components are \(\displaystyle 2\) and \(\displaystyle 6\), and \(\displaystyle \vec{b}\) is not even a scalar multiple of \(\displaystyle \vec{a}\) because \(\displaystyle \tfrac{6}{2}\neq\tfrac{3}{3}\).\(\displaystyle \vec{a}=2\hat{i}+3\hat{j}+6\hat{k}\) and \(\displaystyle \vec{b}=6\hat{i}+3\hat{j}+2\hat{k}\), both of magnitude \(\displaystyle 7\). (Any two unequal vectors of equal magnitude are acceptable.)

  3. Exercise 3

    Write two different vectors having same direction.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    An infinite number of possible answers.
    Two nonzero vectors have the same direction when one is a positive scalar multiple of the other, equivalently when they share the same unit vector \(\displaystyle \hat{a}=\dfrac{\vec{a}}{|\vec{a}|}\).Take \[\vec{a}=\hat{i}+\hat{j}+\hat{k},\qquad \vec{b}=3\hat{i}+3\hat{j}+3\hat{k}=3\vec{a}\] Here \(\displaystyle |\vec{a}|=\sqrt{3}\) and \(\displaystyle |\vec{b}|=\sqrt{9+9+9}=3\sqrt{3}\), so \[\hat{a}=\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})=\frac{3\hat{i}+3\hat{j}+3\hat{k}}{3\sqrt{3}}=\hat{b}\] The multiplier \(\displaystyle 3\) is positive, so the direction is the same (a negative multiplier would reverse it), while the magnitudes \(\displaystyle \sqrt{3}\) and \(\displaystyle 3\sqrt{3}\) differ, so the vectors are different.\(\displaystyle \vec{a}=\hat{i}+\hat{j}+\hat{k}\) and \(\displaystyle \vec{b}=3\hat{i}+3\hat{j}+3\hat{k}\).

  4. Exercise 4

    Find the values of x\displaystyle x and y\displaystyle y so that the vectors 2i^+3j^\displaystyle 2 \hat{i}+3 \hat{j} and xi^+yj^\displaystyle x \hat{i}+y \hat{j} are equal.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle x=2, y=3\)
    Equality of vectors is componentwise: two vectors are equal only if their corresponding components are equal. \[2\hat{i}+3\hat{j}=x\hat{i}+y\hat{j}\] Comparing the coefficients of \(\displaystyle \hat{i}\) and of \(\displaystyle \hat{j}\) separately, \[x=2,\qquad y=3\]\(\displaystyle x=2\) and \(\displaystyle y=3\).

  5. Exercise 5

    Find the scalar and vector components of the vector with initial point (2,1)\displaystyle (2,1) and terminal point (−5,7)\displaystyle (-5, 7).

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    -$\displaystyle 7$ and \(\displaystyle 6 ;-7 \hat{i}\) and \(\displaystyle 6 \hat{j}\)
    For initial point A\(\displaystyle (x_{1},y_{1})\) and terminal point B\(\displaystyle (x_{2},y_{2})\), \[\overrightarrow{\mathrm{AB}}=(x_{2}-x_{1})\hat{i}+(y_{2}-y_{1})\hat{j}\] terminal minus initial, in that order (reversing the order would flip both signs).With A\(\displaystyle (2,1)\) and B\(\displaystyle (-5,7)\), \[\overrightarrow{\mathrm{AB}}=(-5-2)\hat{i}+(7-1)\hat{j}=-7\hat{i}+6\hat{j}\]Scalar components: \(\displaystyle -7\) and \(\displaystyle 6\). Vector components: \(\displaystyle -7\hat{i}\) and \(\displaystyle 6\hat{j}\).

  6. Exercise 6

    Find the sum of the vectors a⃗=i^−2j^+k^,b⃗=−2i^+4j^+5k^\displaystyle \vec{a}=\hat{i}-2 \hat{j}+\hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k} and c⃗=i^−6j^−7k^\displaystyle \vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle -4 \hat{j}-\hat{k}\)
    Vectors are added componentwise: collect the \(\displaystyle \hat{i}\), \(\displaystyle \hat{j}\) and \(\displaystyle \hat{k}\) coefficients separately. \[\vec{a}+\vec{b}+\vec{c}=(1-2+1)\hat{i}+(-2+4-6)\hat{j}+(1+5-7)\hat{k}\] \[=0\,\hat{i}-4\hat{j}-\hat{k}\]\(\displaystyle \vec{a}+\vec{b}+\vec{c}=-4\hat{j}-\hat{k}\).

  7. Exercise 7

    Find the unit vector in the direction of the vector a⃗=i^+j^+2k^\displaystyle \vec{a}=\hat{i}+\hat{j}+2 \hat{k}.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle \frac{1}{\sqrt{6}} \hat{i}+\frac{1}{\sqrt{6}} \hat{j}+\frac{2}{\sqrt{6}} \hat{k}\)
    The unit vector in the direction of \(\displaystyle \vec{a}\) is \(\displaystyle \hat{a}=\dfrac{\vec{a}}{|\vec{a}|}\), defined for \(\displaystyle \vec{a}\neq\vec{0}\). \[|\vec{a}|=\sqrt{1^{2}+1^{2}+2^{2}}=\sqrt{6}\] \[\hat{a}=\frac{1}{\sqrt{6}}(\hat{i}+\hat{j}+2\hat{k})=\frac{1}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}+\frac{2}{\sqrt{6}}\hat{k}\] Check: \(\displaystyle \frac{1}{6}+\frac{1}{6}+\frac{4}{6}=1\).\(\displaystyle \hat{a}=\frac{1}{\sqrt{6}}(\hat{i}+\hat{j}+2\hat{k})\).

  8. Exercise 8

    Find the unit vector in the direction of vector PQ→\displaystyle \overrightarrow{\mathrm{PQ}}, where P and Q are the points (1\displaystyle 1, 2\displaystyle 2, 3\displaystyle 3) and (4\displaystyle 4, 5\displaystyle 5, 6\displaystyle 6), respectively.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle \frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}\)
    \(\displaystyle \overrightarrow{\mathrm{PQ}}\) = position vector of Q \(\displaystyle -\) position vector of P. \[\overrightarrow{\mathrm{PQ}}=(4-1)\hat{i}+(5-2)\hat{j}+(6-3)\hat{k}=3\hat{i}+3\hat{j}+3\hat{k}\] \[|\overrightarrow{\mathrm{PQ}}|=\sqrt{3^{2}+3^{2}+3^{2}}=\sqrt{27}=3\sqrt{3}\] Divide the vector by its magnitude: \[\widehat{\mathrm{PQ}}=\frac{3\hat{i}+3\hat{j}+3\hat{k}}{3\sqrt{3}}=\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\]Unit vector along \(\displaystyle \overrightarrow{\mathrm{PQ}}\) is \(\displaystyle \frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}\).

  9. Exercise 9

    For given vectors, a⃗=2i^−j^+2k^\displaystyle \vec{a}=2 \hat{i}-\hat{j}+2 \hat{k} and b⃗=−i^+j^−k^\displaystyle \vec{b}=-\hat{i}+\hat{j}-\hat{k}, find the unit vector in the direction of the vector a⃗+b⃗\displaystyle \vec{a}+\vec{b}.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle \frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}\)
    Add first, then normalise — the unit vector is required along \(\displaystyle \vec{a}+\vec{b}\), not along \(\displaystyle \vec{a}\) or \(\displaystyle \vec{b}\). \[\vec{a}+\vec{b}=(2-1)\hat{i}+(-1+1)\hat{j}+(2-1)\hat{k}=\hat{i}+0\,\hat{j}+\hat{k}\] \[|\vec{a}+\vec{b}|=\sqrt{1^{2}+0^{2}+1^{2}}=\sqrt{2}\] \[\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{1}{\sqrt{2}}(\hat{i}+\hat{k})\]Required unit vector \(\displaystyle =\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}\).

  10. Exercise 10

    Find a vector in the direction of vector 5i^−j^+2k^\displaystyle 5 \hat{i}-\hat{j}+2 \hat{k} which has magnitude 8\displaystyle 8 units.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle \frac{40}{\sqrt{30}} \hat{i}-\frac{8}{\sqrt{30}} \hat{j}+\frac{16}{\sqrt{30}} \hat{k}\)
    A vector of magnitude \(\displaystyle m\) in the direction of \(\displaystyle \vec{a}\) is \(\displaystyle m\,\hat{a}=m\dfrac{\vec{a}}{|\vec{a}|}\); scaling a unit vector by \(\displaystyle m\) is what fixes the length. \[\vec{a}=5\hat{i}-\hat{j}+2\hat{k},\qquad |\vec{a}|=\sqrt{25+1+4}=\sqrt{30}\] \[8\hat{a}=\frac{8}{\sqrt{30}}(5\hat{i}-\hat{j}+2\hat{k})=\frac{40}{\sqrt{30}}\hat{i}-\frac{8}{\sqrt{30}}\hat{j}+\frac{16}{\sqrt{30}}\hat{k}\] Check: its magnitude is \(\displaystyle \frac{8}{\sqrt{30}}\cdot\sqrt{30}=8\).Required vector \(\displaystyle =\frac{8}{\sqrt{30}}\left(5\hat{i}-\hat{j}+2\hat{k}\right)\).