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NCERT Solutions · Class 12 Mathematics Vector Algebra

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EXERCISE 10.1 1–5 (part 1 of 8)

  1. Exercise 1

    Represent graphically a displacement of 40 km,30∘\displaystyle 40 \mathrm{~km}, 30^{\circ} east of north.

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    NCERT’s answer
    In the adjoining figure, the vector \(\displaystyle \overrightarrow{\mathrm{OP}}\) represents the required displacement. ![](https://cdn.mathpix.com/cropped/476781f9-a560-$\displaystyle 4705$-81ba-98e12b199800-13.jpg?height=$\displaystyle 441$&width=$\displaystyle 448$&top_left_y=$\displaystyle 977$&top_left_x=$\displaystyle 529$)
    NCERT_Solution_Class12_Maths_Ch10_Ex10-1_Q1A displacement is represented graphically by a directed line segment: its length, on a stated scale, gives the magnitude, and the direction of the arrow gives the direction of the displacement.Choose a scale first, otherwise "length" is meaningless. Take \[1\ \mathrm{cm} \equiv 10\ \mathrm{km}, \qquad\text{so}\qquad 40\ \mathrm{km}\ \equiv\ 4\ \mathrm{cm}. \]Now fix the direction. The phrase \(\displaystyle 30^{\circ}\) east of north is measured starting from the north direction and turning through \(\displaystyle 30^{\circ}\) towards the east — it is not \(\displaystyle 30^{\circ}\) measured from the east, and not \(\displaystyle 30^{\circ}\) from the horizontal. This is the step most often got wrong.So, from a starting point O draw the north line ON. Rotate ON through \(\displaystyle 30^{\circ}\) towards OE (east) and along this ray mark P with \(\displaystyle \mathrm{OP}=4\ \mathrm{cm}\), putting the arrow head at P.The vector \(\displaystyle \overrightarrow{\mathrm{OP}}\) so drawn represents the required displacement: \(\displaystyle |\overrightarrow{\mathrm{OP}}|=40\ \mathrm{km}\), directed at \(\displaystyle 30^{\circ}\) east of north (equivalently, making \(\displaystyle 60^{\circ}\) with the east direction).

  2. Exercise 2

    Classify the following measures as scalars and vectors.
    (i)
    10\displaystyle 10 kg
    (ii)
    2\displaystyle 2 meters north-west
    (iii)
    40\displaystyle 40°
    (iv)
    40\displaystyle 40 watt
    (v)
    10−19\displaystyle 10^{-19} coulomb
    (vi)
    20 m/s2\displaystyle 20 \mathrm{~m} / \mathrm{s}^{2}

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    (i)
    scalar (ii) vector (iii) scalar (iv) scalar (v) scalar (vi) vector
    Test used: a quantity that is completely specified by a magnitude (a number with its unit) alone is a scalar; a quantity that needs a magnitude and a direction, and that combines by the triangle law of addition, is a vector.
    (i)
    \(\displaystyle 10\ \mathrm{kg}\) — this is a mass. Only a number and a unit are given, and mass has no direction. Scalar.
    (ii)
    \(\displaystyle 2\) metres north-west — this is a displacement (or distance in a stated direction). The words "north-west" are part of the answer, so a direction is essential. Vector.
    (iii)
    \(\displaystyle 40^{\circ}\) — a measure of an angle. An angle is a pure magnitude; the plane angle itself does not point anywhere. Scalar.
    (iv)
    \(\displaystyle 40\ \mathrm{watt}\) — this is power. Fully described by its magnitude. Scalar.
    (v)
    \(\displaystyle 10^{-19}\) coulomb — electric charge. Charge has sign but not direction, and a sign is not a direction. Scalar.
    (vi)
    \(\displaystyle 20\ \mathrm{m}/\mathrm{s}^{2}\) — this is an acceleration, the rate of change of velocity; velocity is directed, hence so is its rate of change. Vector.
    Final answer: scalars — (i), (iii), (iv), (v); vectors — (ii), (vi).

  3. Exercise 3

    Classify the following as scalar and vector quantities.
    (i)
    time period
    (ii)
    distance
    (iii)
    force
    (iv)
    velocity
    (v)
    work done

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    NCERT’s answer
    (i)
    scalar (ii) scalar (iii) vector (iv) vector (v) scalar
    Same test: a scalar is fixed by magnitude alone; a vector needs magnitude together with a direction.
    (i)
    Time period — an interval of time, specified by a number of seconds alone. Scalar.
    (ii)
    Distance — the length of the path actually covered. It carries no direction (this is exactly what separates distance from displacement, which is a vector). Scalar.
    (iii)
    Force — a push or pull, and the line and sense along which it acts must be stated; forces add by the triangle/parallelogram law. Vector.
    (iv)
    Velocity — displacement per unit time, so it inherits the direction of the displacement (again, contrast speed, which is a scalar). Vector.
    (v)
    Work done — obtained as \(\displaystyle W=\vec{F}\cdot\vec{d}\), a scalar product of two vectors, and a scalar product is a number. Scalar.
    Final answer: scalars — (i) time period, (ii) distance, (v) work done; vectors — (iii) force, (iv) velocity.

  4. Exercise 4

    NCERT_Question_Class12_Maths_Ch10_Ex10-1_Q4
    In Fig 10.6\displaystyle 10.6 (a square), identify the following vectors.
    (i)
    Coinitial
    (ii)
    Equal
    (iii)
    Collinear but not equal

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    (i)
    Vectors \(\displaystyle \vec{a}\) and \(\displaystyle \vec{b}\) are coinitial (ii) Vectors \(\displaystyle \vec{b}\) and \(\displaystyle \vec{d}\) are equal (iii) Vectors \(\displaystyle \vec{a}\) and \(\displaystyle \vec{c}\) are collinear but not equal
    Read the arrowheads in Fig 10.6. Writing the square's corners as \(\displaystyle P\) (top-left), \(\displaystyle Q\) (top-right), \(\displaystyle R\) (bottom-right) and \(\displaystyle S\) (bottom-left):\[\vec{a}=\overrightarrow{PQ},\qquad \vec{b}=\overrightarrow{QR},\qquad \vec{c}=\overrightarrow{RS},\qquad \vec{d}=\overrightarrow{PS}\](i) Coinitial. Coinitial vectors share an initial point. The initial points are \(\displaystyle P,\;Q,\;R,\;P\) respectively, so the pair that begins at the same corner is \(\displaystyle \vec{a}\) and \(\displaystyle \vec{d}\), both starting at \(\displaystyle P\).(ii) Equal. Equal vectors have the same magnitude AND the same direction, wherever they are drawn. \(\displaystyle \vec{b}\) runs down the right side and \(\displaystyle \vec{d}\) runs down the left side; both have length equal to the side of the square and both point in the same direction, so \(\displaystyle \vec{b}=\vec{d}\).(iii) Collinear but not equal. \(\displaystyle \vec{a}\) and \(\displaystyle \vec{c}\) are both horizontal, so they are parallel and hence collinear; they have the same magnitude but point in OPPOSITE directions, so they are not equal. In fact \(\displaystyle \vec{c}=-\vec{a}\).

  5. Exercise 5

    Answer the following as true or false.
    (i)
    a⃗\displaystyle \vec{a} and −a⃗\displaystyle -\vec{a} are collinear.
    (ii)
    Two collinear vectors are always equal in magnitude.
    (iii)
    Two vectors having same magnitude are collinear.
    (iv)
    Two collinear vectors having the same magnitude are equal.

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    NCERT’s answer
    (i)
    True (ii) False (iii) False (iv) False
    Definitions being tested: two (non-zero) vectors are collinear (parallel) if each is a scalar multiple of the other, i.e. they are parallel to one and the same line, whatever their magnitudes and whichever way they point; two vectors are equal only if they have the same magnitude and the same direction.
    (i)
    \(\displaystyle \vec{a}\) and \(\displaystyle -\vec{a}\) are collinear. TRUE. Since \(\displaystyle -\vec{a}=(-1)\vec{a}\), it is a scalar multiple of \(\displaystyle \vec{a}\), so it is parallel to the same line (pointing the opposite way, which collinearity permits).
    (ii)
    Two collinear vectors are always equal in magnitude. FALSE. Collinearity restricts direction only, not length. Counterexample: \(\displaystyle \vec{a}\) and \(\displaystyle 2\vec{a}\) are collinear, but \(\displaystyle |2\vec{a}|=2|\vec{a}|\neq|\vec{a}|\) for any \(\displaystyle \vec{a}\neq\vec{0}\).
    (iii)
    Two vectors having same magnitude are collinear. FALSE. Equality of magnitudes says nothing about direction. Counterexample: \(\displaystyle \hat{i}\) and \(\displaystyle \hat{j}\) both have magnitude \(\displaystyle 1\), yet they are perpendicular, so not parallel to one line.
    (iv)
    Two collinear vectors having the same magnitude are equal. FALSE. Collinear leaves two possible senses along the line — same or opposite. Counterexample: \(\displaystyle \vec{a}\) and \(\displaystyle -\vec{a}\) (with \(\displaystyle \vec{a}\neq\vec{0}\)) are collinear and \(\displaystyle |-\vec{a}|=|\vec{a}|\), but their directions are opposite, so \(\displaystyle \vec{a}\neq-\vec{a}\).
    Final answer: (i) True; (ii) False; (iii) False; (iv) False.