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NCERT Exemplar · Class 12 Mathematics Vector Algebra

45 questions · 45 still being checked

EXERCISE 10.3 21–30 (part 3 of 5)

  1. Choose the correct answer from the given four options in each of the Exercises from $\displaystyle 19$ to $\displaystyle 33$ (M.C.Q)

    Exercise 21

    The vector having initial and terminal points as (2,5,0)\displaystyle (2,5,0) and (−3,7,4)\displaystyle (-3,7,4), respectively is
    (A)
    −i^+12j^+4k^\displaystyle -\hat{i}+12 \hat{j}+4 \hat{k}
    (B)
    5i^+2j^−4k^\displaystyle 5 \hat{i}+2 \hat{j}-4 \hat{k}
    (C)
    −5i^+2j^+4k^\displaystyle -5 \hat{i}+2 \hat{j}+4 \hat{k}
    (D)
    i^+j^+k^\displaystyle \hat{i}+\hat{j}+\hat{k}

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    NCERT’s answer
    C
    (C) \(\displaystyle -5\hat i+2\hat j+4\hat k\)Terminal point minus initial point:\[(-3-2)\hat i+(7-5)\hat j+(4-0)\hat k = -5\hat i+2\hat j+4\hat k \]

  2. Exercise 22

    The angle between two vectors a⃗\displaystyle \vec{a} and b⃗\displaystyle \vec{b} with magnitudes 3\displaystyle \sqrt{3} and 4\displaystyle 4, respectively, and a⃗.b⃗=23\displaystyle \vec{a} . \vec{b}=2 \sqrt{3} is
    (A)
    π6\displaystyle \frac{\pi}{6}
    (B)
    π3\displaystyle \frac{\pi}{3}
    (C)
    π2\displaystyle \frac{\pi}{2}
    (D)
    5π2\displaystyle \frac{5 \pi}{2}

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    NCERT’s answer
    B
    (B) \(\displaystyle \dfrac{\pi}{3}\)\[\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} = \frac{2\sqrt3}{\sqrt3\cdot 4} = \frac12 \]\[\theta = \frac{\pi}{3} \]

  3. Exercise 23

    Find the value of λ\displaystyle \lambda such that the vectors a⃗=2i^+λj^+k^\displaystyle \vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k} and b⃗=i^+2j^+3k^\displaystyle \vec{b}=\hat{i}+2 \hat{j}+3 \hat{k} are orthogonal
    (A)
    0\displaystyle 0 (B) 1\displaystyle 1 (C) 32\displaystyle \frac{3}{2}
    (D)
    −52\displaystyle -\frac{5}{2}

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    NCERT’s answer
    D
    (D) \(\displaystyle -\dfrac52\)Orthogonal vectors have zero dot product:\[\vec a\cdot\vec b = 2(1)+\lambda(2)+1(3) = 0 \]\[2\lambda+5=0 \Rightarrow \lambda=-\frac52 \]

  4. Exercise 24

    The value of λ\displaystyle \lambda for which the vectors 3i^−6j^+k^\displaystyle 3 \hat{i}-6 \hat{j}+\hat{k} and 2i^−4j^+λk^\displaystyle 2 \hat{i}-4 \hat{j}+\lambda \hat{k} are parallel is
    (A)
    23\displaystyle \frac{2}{3}
    (B)
    32\displaystyle \frac{3}{2}
    (C)
    52\displaystyle \frac{5}{2}
    (D)
    25\displaystyle \frac{2}{5}

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    NCERT’s answer
    A
    (A) \(\displaystyle \dfrac23\)Parallel vectors have proportional components:\[\frac{3}{2} = \frac{-6}{-4} = \frac{1}{\lambda} \]\[\lambda = \frac23 \]

  5. Exercise 25

    The vectors from origin to the points A and B are a⃗=2i^−3j^+2k^\displaystyle \vec{a}=2 \hat{i}-3 \hat{j}+2 \hat{k} and b⃗=2i^+3j^+k^\displaystyle \vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}, respectively, then the area of triangle OAB is
    (A)
    340\displaystyle 340
    (B)
    25\displaystyle \sqrt{25}
    (C)
    229\displaystyle \sqrt{229}
    (D)
    12229\displaystyle \frac{1}{2} \sqrt{229}

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    NCERT’s answer
    D
    (D) \(\displaystyle \tfrac{1}{2}\sqrt{229}\)\[\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ 2&-3&2\\ 2&3&1\end{vmatrix}=-9\hat{i}+2\hat{j}+12\hat{k} \]\[|\vec{a}\times\vec{b}|=\sqrt{81+4+144}=\sqrt{229} \]\[\text{Area of } \triangle OAB=\tfrac{1}{2}|\vec{a}\times\vec{b}|=\tfrac{1}{2}\sqrt{229} \]

  6. Exercise 26

    For any vector a⃗\displaystyle \vec{a}, the value of (a⃗×i^)2+(a⃗×j^)2+(a⃗×k^)2\displaystyle (\vec{a} \times \hat{i})^2+(\vec{a} \times \hat{j})^2+(\vec{a} \times \hat{k})^2 is equal to
    (A)
    a⃗2\displaystyle \vec{a}^2
    (B)
    3a⃗2\displaystyle 3 \vec{a}^2
    (C)
    4a⃗2\displaystyle 4 \vec{a}^2
    (D)
    2a⃗2\displaystyle 2 \vec{a}^2

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    NCERT’s answer
    D
    (D) \(\displaystyle 2\vec{a}^2\)For a unit vector \(\displaystyle \hat{u}\):\[(\vec{a}\times\hat{u})^2=|\vec{a}|^2|\hat{u}|^2-(\vec{a}\cdot\hat{u})^2=|\vec{a}|^2-(\vec{a}\cdot\hat{u})^2 \]\[\text{Sum}=3|\vec{a}|^2-\left[(\vec{a}\cdot\hat{i})^2+(\vec{a}\cdot\hat{j})^2+(\vec{a}\cdot\hat{k})^2\right] \]\[(\vec{a}\cdot\hat{i})^2+(\vec{a}\cdot\hat{j})^2+(\vec{a}\cdot\hat{k})^2=a_1^2+a_2^2+a_3^2=|\vec{a}|^2 \]\[\text{Sum}=3|\vec{a}|^2-|\vec{a}|^2=2\vec{a}^2 \]

  7. Exercise 27

    If ∣a⃗∣=10,∣b⃗∣=2\displaystyle |\vec{a}|=10,|\vec{b}|=2 and a⃗⋅b⃗=12\displaystyle \vec{a} \cdot \vec{b}=12, then value of ∣a⃗×b⃗∣\displaystyle |\vec{a} \times \vec{b}| is
    (A)
    5\displaystyle 5 (B) 10\displaystyle 10
    (C)
    14\displaystyle 14
    (D)
    16\displaystyle 16

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    D
    (D) \(\displaystyle 16\)\[|\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2=|\vec{a}|^2|\vec{b}|^2 \]\[|\vec{a}\times\vec{b}|^2=10^2\cdot 2^2-12^2=400-144=256 \]\[|\vec{a}\times\vec{b}|=16 \]

  8. Exercise 28

    The vectors λi^+j^+2k^,i^+λj^−k^\displaystyle \lambda \hat{i}+\hat{j}+2 \hat{k}, \hat{i}+\lambda \hat{j}-\hat{k} and 2i^−j^+λk^\displaystyle 2 \hat{i}-\hat{j}+\lambda \hat{k} are coplanar if
    (A)
    λ=−2\displaystyle \lambda=-2
    (B)
    λ=0\displaystyle \lambda=0
    (C)
    λ=1\displaystyle \lambda=1
    (D)
    λ=−1\displaystyle \lambda=-1

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    NCERT’s answer
    A
    (A) \(\displaystyle \lambda=-2\)Coplanar means the scalar triple product is zero.\[\begin{vmatrix}\lambda&1&2\\ 1&\lambda&-1\\ 2&-1&\lambda\end{vmatrix}=\lambda(\lambda^2-1)-(\lambda+2)+2(-1-2\lambda)=0 \]\[\lambda^3-6\lambda-4=0 \]\[(\lambda+2)(\lambda^2-2\lambda-2)=0 \]The quadratic gives \(\displaystyle \lambda=1\pm\sqrt{3}\), not among the options, so \(\displaystyle \lambda=-2\).

  9. Exercise 29

    If a⃗,b⃗,c⃗\displaystyle \vec{a}, \vec{b}, \vec{c} are unit vectors such that a⃗+b⃗+c⃗=0→\displaystyle \vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}, then the value of a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗\displaystyle \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} is
    (A)
    1\displaystyle 1 (B) 3\displaystyle 3 (C) −32\displaystyle -\frac{3}{2}
    (D)
    None of these

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    NCERT’s answer
    C
    (C) \(\displaystyle -\tfrac32\)\[|\vec a+\vec b+\vec c|^2=0 \]\[|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0 \]\[1+1+1+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0 \]\[\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a=-\tfrac32 \]

  10. Exercise 30

    Projection vector of a⃗\displaystyle \vec{a} on b⃗\displaystyle \vec{b} is
    (A)
    (a⃗⋅b⃗∣b⃗∣2)b⃗\displaystyle \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b}
    (B)
    a⃗⋅b⃗∣b⃗∣\displaystyle \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}
    (C)
    a⃗⋅b⃗∣a⃗∣\displaystyle \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}
    (D)
    (a⃗⋅b⃗∣a⃗∣2)b^\displaystyle \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right) \hat{b}

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    NCERT’s answer
    A
    (A) \(\displaystyle \left(\dfrac{\vec a\cdot\vec b}{|\vec b|^{2}}\right)\vec b\)NCERT_Solution_Class12_Maths_Exemplar_Ch10_Ex10-3_Q30With \(\displaystyle \theta\) the angle between \(\displaystyle \vec a\) and \(\displaystyle \vec b\), the projection vector is \(\displaystyle OP\) along the unit vector of \(\displaystyle \vec b\):\[OP=|\vec a|\cos\theta=\frac{\vec a\cdot\vec b}{|\vec b|} \]\[\hat b=\frac{\vec b}{|\vec b|} \]\[\overrightarrow{OP}=\frac{\vec a\cdot\vec b}{|\vec b|}\cdot\frac{\vec b}{|\vec b|}=\frac{\vec a\cdot\vec b}{|\vec b|^{2}}\,\vec b \]