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NCERT Exemplar · Class 12 Mathematics Vector Algebra

45 questions · 45 still being checked

EXERCISE 10.3 11–20 (part 2 of 5)

  1. Exercise 11

    Find the sine of the angle between the vectors a⃗=3i^+j^+2k^\displaystyle \vec{a}=3 \hat{i}+\hat{j}+2 \hat{k} and b⃗=2i^−2j^+4k^\displaystyle \vec{b}=2 \hat{i}-2 \hat{j}+4 \hat{k}.

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    NCERT’s answer
    \(\displaystyle \frac{2}{\sqrt{7}}\)
    \[\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\3&1&2\\2&-2&4\end{vmatrix}=8\hat i-8\hat j-8\hat k \]\[|\vec a\times\vec b|=\sqrt{64+64+64}=8\sqrt3 \]\[|\vec a|=\sqrt{9+1+4}=\sqrt{14},\qquad |\vec b|=\sqrt{4+4+16}=\sqrt{24} \]\[\sin\theta=\frac{|\vec a\times\vec b|}{|\vec a|\,|\vec b|}=\frac{8\sqrt3}{\sqrt{14}\,\sqrt{24}}=\frac{8\sqrt3}{4\sqrt{21}}=\frac{2}{\sqrt7} \]Answer: \(\displaystyle \sin\theta=\dfrac{2}{\sqrt7}\)

  2. Exercise 12

    If A,B,C,D\displaystyle \mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D} are the points with position vectors i^+j^−k^,2i^−j^+3k^\displaystyle \hat{i}+\hat{j}-\hat{k}, 2 \hat{i}-\hat{j}+3 \hat{k}, 2i^−3k^,3i^−2j^+k^\displaystyle 2 \hat{i}-3 \hat{k}, 3 \hat{i}-2 \hat{j}+\hat{k}, respectively, find the projection of AB→\displaystyle \overrightarrow{\mathrm{AB}} along CD→\displaystyle \overrightarrow{\mathrm{CD}}.

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    NCERT’s answer
    \(\displaystyle \sqrt{21}\)
    \[\overrightarrow{AB}=(2-1)\hat i+(-1-1)\hat j+(3+1)\hat k=\hat i-2\hat j+4\hat k \]\[\overrightarrow{CD}=(3-2)\hat i+(-2-0)\hat j+(1+3)\hat k=\hat i-2\hat j+4\hat k \]\[\text{projection of }\overrightarrow{AB}\text{ along }\overrightarrow{CD}=\frac{\overrightarrow{AB}\cdot\overrightarrow{CD}}{|\overrightarrow{CD}|} \]\[=\frac{1+4+16}{\sqrt{1+4+16}}=\frac{21}{\sqrt{21}}=\sqrt{21} \]Answer: \(\displaystyle \sqrt{21}\) units

  3. Exercise 13

    Using vectors, find the area of the triangle ABC with vertices A(1,2,3)\displaystyle \mathrm{A}(1,2,3), B(2\displaystyle 2, -1\displaystyle 1, 4\displaystyle 4) and C(4\displaystyle 4, 5\displaystyle 5, - 1\displaystyle 1).

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    NCERT’s answer
    \(\displaystyle \frac{\sqrt{274}}{2}\)
    \[\overrightarrow{AB}=(2-1)\hat i+(-1-2)\hat j+(4-3)\hat k=\hat i-3\hat j+\hat k \]\[\overrightarrow{AC}=(4-1)\hat i+(5-2)\hat j+(-1-3)\hat k=3\hat i+3\hat j-4\hat k \]NCERT_Solution_Class12_Maths_Exemplar_Ch10_Ex10-3_Q13\[\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-3&1\\3&3&-4\end{vmatrix}=9\hat i+7\hat j+12\hat k \]\[|\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{81+49+144}=\sqrt{274} \]\[\text{Area}=\tfrac12\,|\overrightarrow{AB}\times\overrightarrow{AC}|=\tfrac12\sqrt{274} \]Answer: \(\displaystyle \tfrac12\sqrt{274}\) square units

  4. Exercise 14

    Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.

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    Let \(\displaystyle ABCD\) and \(\displaystyle ABEF\) stand on the base \(\displaystyle AB\), with \(\displaystyle D, F, C, E\) on a line parallel to \(\displaystyle AB\).NCERT_Solution_Class12_Maths_Exemplar_Ch10_Ex10-3_Q14\[\overrightarrow{AB}=\vec a,\qquad \overrightarrow{AD}=\vec b \]\[\text{Area}(ABCD)=|\vec a\times\vec b| \]\(\displaystyle F\) lies on the line through \(\displaystyle D\) parallel to \(\displaystyle \vec a\):\[\overrightarrow{AF}=\overrightarrow{AD}+\overrightarrow{DF}=\vec b+\lambda\vec a \]\[\text{Area}(ABEF)=|\vec a\times\overrightarrow{AF}|=|\vec a\times(\vec b+\lambda\vec a)| \]\[=|\vec a\times\vec b+\lambda(\vec a\times\vec a)|=|\vec a\times\vec b| \qquad (\vec a\times\vec a=\vec 0) \]Answer: \(\displaystyle \text{Area}(ABEF)=\text{Area}(ABCD)=|\vec a\times\vec b|\)

  5. Exercise 15

    Prove that in any triangle ABC,cos⁡A=b2+c2−a22bc\displaystyle \mathrm{ABC}, \cos \mathrm{A}=\frac{b^2+c^2-a^2}{2 b c}, where a,b,c\displaystyle a, b, c are the magnitudes of the sides opposite to the vertices A,B,C\displaystyle \mathrm{A}, \mathrm{B}, \mathrm{C}, respectively.

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    In \(\displaystyle \triangle ABC\), \(\displaystyle |\overrightarrow{BC}|=a\), \(\displaystyle |\overrightarrow{AC}|=b\), \(\displaystyle |\overrightarrow{AB}|=c\), and the angle between \(\displaystyle \overrightarrow{AB}\) and \(\displaystyle \overrightarrow{AC}\) is \(\displaystyle A\).NCERT_Solution_Class12_Maths_Exemplar_Ch10_Ex10-3_Q15\[\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB} \]\[a^2=\overrightarrow{BC}\cdot\overrightarrow{BC}=(\overrightarrow{AC}-\overrightarrow{AB})\cdot(\overrightarrow{AC}-\overrightarrow{AB}) \]\[a^2=|\overrightarrow{AC}|^2+|\overrightarrow{AB}|^2-2\,\overrightarrow{AB}\cdot\overrightarrow{AC} \]\[\overrightarrow{AB}\cdot\overrightarrow{AC}=bc\cos A \]\[a^2=b^2+c^2-2bc\cos A \]\[\cos A=\frac{b^2+c^2-a^2}{2bc} \]Answer: \(\displaystyle \cos A=\dfrac{b^2+c^2-a^2}{2bc}\)

  6. Exercise 16

    If a⃗,b⃗,c⃗\displaystyle \vec{a}, \vec{b}, \vec{c} determine the vertices of a triangle, show that 12[b⃗×c⃗+c⃗×a⃗+a⃗×b⃗]\displaystyle \frac{1}{2}[\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}] gives the vector area of the triangle. Hence deduce the condition that the three points a⃗,b⃗,c⃗\displaystyle \vec{a}, \vec{b}, \vec{c} are collinear. Also find the unit vector normal to the plane of the triangle.

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    NCERT’s answer
    \(\displaystyle \hat{n}=\frac{\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}}{|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|}\)
    Let \(\displaystyle A, B, C\) have position vectors \(\displaystyle \vec a, \vec b, \vec c\).NCERT_Solution_Class12_Maths_Exemplar_Ch10_Ex10-3_Q16\[\overrightarrow{AB}=\vec b-\vec a,\qquad \overrightarrow{AC}=\vec c-\vec a \]\[\text{Vector area}=\tfrac12\,\overrightarrow{AB}\times\overrightarrow{AC}=\tfrac12(\vec b-\vec a)\times(\vec c-\vec a) \]\[=\tfrac12\left[\vec b\times\vec c-\vec b\times\vec a-\vec a\times\vec c+\vec a\times\vec a\right] \]\[=\tfrac12\left[\vec b\times\vec c+\vec c\times\vec a+\vec a\times\vec b\right] \]Collinear points enclose zero area:\[\vec a\times\vec b+\vec b\times\vec c+\vec c\times\vec a=\vec 0 \]The vector area is perpendicular to the plane of the triangle:\[\hat n=\frac{\vec a\times\vec b+\vec b\times\vec c+\vec c\times\vec a}{\left|\vec a\times\vec b+\vec b\times\vec c+\vec c\times\vec a\right|} \]Answer: The vector area is \(\displaystyle \tfrac12[\vec b\times\vec c+\vec c\times\vec a+\vec a\times\vec b]\); the points are collinear when \(\displaystyle \vec a\times\vec b+\vec b\times\vec c+\vec c\times\vec a=\vec 0\); the unit normal is \(\displaystyle \hat n\) above.

  7. Exercise 17

    Show that area of the parallelogram whose diagonals are given by a⃗\displaystyle \vec{a} and b⃗\displaystyle \vec{b} is ∣a⃗×b⃗∣2\displaystyle \frac{|\vec{a} \times \vec{b}|}{2}. Also find the area of the parallelogram whose diagonals are 2i^−j^+k^\displaystyle 2 \hat{i}-\hat{j}+\hat{k} and i^+3j^−k^\displaystyle \hat{i}+3 \hat{j}-\hat{k}.

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    NCERT’s answer
    \(\displaystyle \frac{\sqrt{62}}{2}\)
    Take adjacent sides \(\displaystyle \vec{AB}=\vec p\), \(\displaystyle \vec{AD}=\vec q\), with diagonals \(\displaystyle \vec{AC}=\vec a\), \(\displaystyle \vec{BD}=\vec b\).NCERT_Solution_Class12_Maths_Exemplar_Ch10_Ex10-3_Q17\[\vec a = \vec p + \vec q \qquad \vec b = \vec q - \vec p \]\[\vec a\times\vec b = (\vec p+\vec q)\times(\vec q-\vec p) \]\[= \vec p\times\vec q - \vec p\times\vec p + \vec q\times\vec q - \vec q\times\vec p \]\[= \vec p\times\vec q + \vec p\times\vec q = 2\,(\vec p\times\vec q) \]\[\text{Area} = |\vec p\times\vec q| = \frac{|\vec a\times\vec b|}{2} \]For \(\displaystyle \vec a = 2\hat i-\hat j+\hat k\), \(\displaystyle \vec b = \hat i+3\hat j-\hat k\):\[\vec a\times\vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{vmatrix} = -2\hat i+3\hat j+7\hat k \]\[|\vec a\times\vec b| = \sqrt{4+9+49} = \sqrt{62} \]\[\text{Area} = \frac{\sqrt{62}}{2} \]Answer: \(\displaystyle \dfrac{\sqrt{62}}{2}\) square units.

  8. Exercise 18

    If a⃗=i^+j^+k^\displaystyle \vec{a}=\hat{i}+\hat{j}+\hat{k} and b⃗=j^−k^\displaystyle \vec{b}=\hat{j}-\hat{k}, find a vector c⃗\displaystyle \vec{c} such that a⃗×c⃗=b⃗\displaystyle \vec{a} \times \vec{c}=\vec{b} and a⃗.c⃗=3\displaystyle \vec{a} . \vec{c}=3.

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    NCERT’s answer
    \(\displaystyle \frac{1}{3}(5 \vec{i}+2 \vec{j}+2 \vec{k})\)
    Let \(\displaystyle \vec c = x\hat i+y\hat j+z\hat k\).\[\vec a\times\vec c = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = (z-y)\hat i+(x-z)\hat j+(y-x)\hat k \]Equate with \(\displaystyle \vec b = 0\,\hat i+\hat j-\hat k\):\[z-y=0, \qquad x-z=1, \qquad y-x=-1 \]\[y=z, \qquad x=z+1 \]\[\vec a\cdot\vec c = x+y+z = 3 \]\[(z+1)+z+z=3 \Rightarrow z=\frac23 \]\[x=\frac53, \qquad y=\frac23 \]Answer: \(\displaystyle \vec c = \dfrac53\hat i+\dfrac23\hat j+\dfrac23\hat k\)
  9. Choose the correct answer from the given four options in each of the Exercises from $\displaystyle 19$ to $\displaystyle 33$ (M.C.Q)

    Exercise 19

    The vector in the direction of the vector i^−2j^+2k^\displaystyle \hat{i}-2 \hat{j}+2 \hat{k} that has magnitude 9\displaystyle 9 is
    (A)
    i^−2j^+2k^\displaystyle \hat{i}-2 \hat{j}+2 \hat{k}
    (B)
    i^−2j^+2k^3\displaystyle \frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3}
    (C)
    3(i^−2j^+2k^)\displaystyle 3(\hat{i}-2 \hat{j}+2 \hat{k})
    (D)
    9(i^−2j^+2k^)\displaystyle 9(\hat{i}-2 \hat{j}+2 \hat{k})

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    NCERT’s answer
    C
    (C) \(\displaystyle 3(\hat i-2\hat j+2\hat k)\)\[|\hat i-2\hat j+2\hat k| = \sqrt{1+4+4} = 3 \]\[\vec r = 9\cdot\frac{\hat i-2\hat j+2\hat k}{3} = 3(\hat i-2\hat j+2\hat k) \]

  10. Exercise 20

    The position vector of the point which divides the join of points 2a⃗−3b⃗\displaystyle 2 \vec{a}-3 \vec{b} and a⃗+b⃗\displaystyle \vec{a}+\vec{b} in the ratio 3\displaystyle 3 : 1\displaystyle 1 is
    (A)
    3a⃗−2b⃗2\displaystyle \frac{3 \vec{a}-2 \vec{b}}{2}
    (B)
    7a⃗−8b⃗4\displaystyle \frac{7 \vec{a}-8 \vec{b}}{4}
    (C)
    3a⃗4\displaystyle \frac{3 \vec{a}}{4}
    (D)
    5a⃗4\displaystyle \frac{5 \vec{a}}{4}

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    NCERT’s answer
    D
    (D) \(\displaystyle \dfrac{5\vec a}{4}\)Section formula, internal division \(\displaystyle 3:1\):\[\vec r = \frac{3(\vec a+\vec b)+1\,(2\vec a-3\vec b)}{3+1} \]\[= \frac{5\vec a}{4} \]