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NCERT Exemplar · Class 12 Mathematics Three Dimensional Geometry

49 questions · 49 still being checked

EXERCISE 11.3 31–40 (part 4 of 5)

  1. Choose the correct answer from the given four options in each of the Exercises from $\displaystyle 29$ to 36.

    Exercise 31

    The distance of the plane r⃗⋅(27i^+37j^−67k^)=1\displaystyle \vec{r} \cdot\left(\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}-\frac{6}{7} \hat{k}\right)=1 from the origin is
    (A)
    1\displaystyle 1 (B) 7\displaystyle 7 (C) 17\displaystyle \frac{1}{7}
    (D)
    None of these

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    A
    (A) \(\displaystyle 1\)For \(\displaystyle \vec r\cdot\vec n=d\), the distance from the origin is \(\displaystyle \dfrac{|d|}{|\vec n|}\).\[|\vec n|=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=\sqrt{\frac{49}{49}}=1 \]\[\text{distance}=\frac{|1|}{1}=1 \]

  2. Exercise 32

    The sine of the angle between the straight line x−23=y−34=z−45\displaystyle \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5} and the plane 2x−2y+z=5\displaystyle 2 x-2 y+z=5 is
    (A)
    1065\displaystyle \frac{10}{6 \sqrt{5}}
    (B)
    452\displaystyle \frac{4}{5 \sqrt{2}}
    (C)
    235\displaystyle \frac{2 \sqrt{3}}{5}
    (D)
    210\displaystyle \frac{\sqrt{2}}{10}

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    D
    (D) \(\displaystyle \dfrac{\sqrt2}{10}\)With \(\displaystyle \vec b\) the direction of the line and \(\displaystyle \vec n\) the normal to the plane:\[\sin\theta=\frac{|\vec b\cdot\vec n|}{|\vec b|\,|\vec n|} \]\[\vec b=3\hat i+4\hat j+5\hat k,\qquad \vec n=2\hat i-2\hat j+\hat k \]\[\vec b\cdot\vec n=6-8+5=3 \]\[|\vec b|=\sqrt{9+16+25}=5\sqrt2,\qquad |\vec n|=\sqrt{4+4+1}=3 \]\[\sin\theta=\frac{3}{15\sqrt2}=\frac{1}{5\sqrt2}=\frac{\sqrt2}{10} \]

  3. Exercise 33

    The reflection of the point (α,β,γ)\displaystyle (\alpha, \beta, \gamma) in the xy\displaystyle x y-plane is
    (A)
    (α,β,0)\displaystyle (\alpha, \beta, 0)
    (B)
    (0,0,γ)\displaystyle (0,0, \gamma)
    (C)
    (−α,−β,γ)\displaystyle (-\alpha,-\beta, \gamma)
    (D)
    (α,β,−γ)\displaystyle (\alpha, \beta,-\gamma)

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    D
    (D) \(\displaystyle (\alpha,\beta,-\gamma)\)The \(\displaystyle xy\)-plane is \(\displaystyle z=0\). For \(\displaystyle Q(\alpha,\beta,-\gamma)\) and \(\displaystyle P(\alpha,\beta,\gamma)\), the midpoint lies on the plane and \(\displaystyle PQ\) is along its normal.\[\text{midpoint of }PQ=\left(\alpha,\ \beta,\ \frac{\gamma-\gamma}{2}\right)=(\alpha,\beta,0) \]\[\overrightarrow{PQ}=-2\gamma\,\hat k \]

  4. Exercise 34

    The area of the quadrilateral ABCD, where A(0,4,1),B(2,3,−1),C(4,5,0)\displaystyle \mathrm{A}(0,4,1), \mathrm{B}(2,3,-1), \mathrm{C}(4,5,0) and D (2\displaystyle 2, 6\displaystyle 6, 2\displaystyle 2), is equal to
    (A)
    9\displaystyle 9 sq. units
    (B)
    18\displaystyle 18 sq. units
    (C)
    27\displaystyle 27 sq. units
    (D)
    81\displaystyle 81 sq. units

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    NCERT’s answer
    A
    (A) \(\displaystyle 9\) sq. units\[\overrightarrow{AB}=2\hat i-\hat j-2\hat k,\qquad \overrightarrow{BC}=2\hat i+2\hat j+\hat k \]\[\overrightarrow{CD}=-2\hat i+\hat j+2\hat k,\qquad \overrightarrow{DA}=-2\hat i-2\hat j-\hat k \]\[\overrightarrow{AB}=-\overrightarrow{CD},\qquad \overrightarrow{BC}=-\overrightarrow{DA} \]So ABCD is a parallelogram, with area \(\displaystyle \left|\overrightarrow{AB}\times\overrightarrow{BC}\right|\).\[\overrightarrow{AB}\times\overrightarrow{BC}=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-1&-2\\2&2&1\end{vmatrix}=3\hat i-6\hat j+6\hat k \]\[\text{Area}=\sqrt{9+36+36}=9 \]

  5. Exercise 35

    The locus represented by xy+yz=0\displaystyle x y+y z=0 is
    (A)
    A pair of perpendicular lines
    (B)
    A pair of parallel lines
    (C)
    A pair of parallel planes
    (D)
    A pair of perpendicular planes

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    NCERT’s answer
    D
    (D) A pair of perpendicular planes\[xy+yz=0 \]\[y\,(x+z)=0 \]\[y=0\quad\text{or}\quad x+z=0 \]These are two planes with normals \(\displaystyle \vec n_1=(0,1,0)\) and \(\displaystyle \vec n_2=(1,0,1)\).\[\vec n_1\cdot\vec n_2=0+0+0=0 \]The normals are perpendicular, so the planes are perpendicular.

  6. Exercise 36

    The plane 2x−3y+6z−11=0\displaystyle 2 x-3 y+6 z-11=0 makes an angle sin⁡−1(α)\displaystyle \sin ^{-1}(\alpha) with x-axis. The value of α\displaystyle \alpha is equal to
    (A)
    32\displaystyle \frac{\sqrt{3}}{2}
    (B)
    23\displaystyle \frac{\sqrt{2}}{3}
    (C)
    27\displaystyle \frac{2}{7}
    (D)
    37\displaystyle \frac{3}{7}

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    NCERT’s answer
    C
    (C) \(\displaystyle \dfrac{2}{7}\)The plane's normal and the x-axis direction \(\displaystyle \hat i\) give the angle with the plane:\[\vec n = 2\hat i - 3\hat j + 6\hat k \]\[|\vec n| = \sqrt{2^2+(-3)^2+6^2} = \sqrt{49} = 7 \]\[\sin\theta = \frac{|\vec n\cdot\hat i|}{|\vec n|\,|\hat i|} = \frac{2}{7} \]\[\alpha = \frac{2}{7} \]
  7. Fill in the blanks in each of the Exercises $\displaystyle 37$ to 41.

    Exercise 37

    A plane passes through the points (2,0,0)(0,3,0)\displaystyle (2,0,0)(0,3,0) and (0,0,4)\displaystyle (0,0,4). The equation of plane is ____.

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    NCERT’s answer
    \(\displaystyle \frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1\)
    \(\displaystyle 6x+4y+3z=12\)The intercepts on the axes are \(\displaystyle a=2,\ b=3,\ c=4\).\[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \]\[\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1 \]\[6x+4y+3z=12 \]

  8. Exercise 38

    The direction cosines of the vector (2i^+2j^−k^)\displaystyle (2 \hat{i}+2 \hat{j}-\hat{k}) are ____.

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    NCERT’s answer
    \(\displaystyle \frac{2}{3}, \frac{2}{3}, \frac{-1}{3}\)
    \(\displaystyle \dfrac{2}{3},\ \dfrac{2}{3},\ -\dfrac{1}{3}\)Divide each component by the magnitude.\[|\vec a| = \sqrt{2^2+2^2+(-1)^2} = 3 \]\[(l,\,m,\,n) = \left(\frac{2}{3},\ \frac{2}{3},\ -\frac{1}{3}\right) \]\[l^2+m^2+n^2 = \frac{4}{9}+\frac{4}{9}+\frac{1}{9} = 1 \]

  9. Exercise 39

    The vector equation of the line x−53=y+47=z−62\displaystyle \frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} is ____.

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    NCERT’s answer
    \(\displaystyle (x-5) \hat{i}+(y+4) \hat{j}+(z-6) \hat{k}=\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})\)
    \(\displaystyle \vec r = (5\hat i-4\hat j+6\hat k)+\lambda(3\hat i+7\hat j+2\hat k)\)Write the line as \(\displaystyle \dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}\).\[\frac{x-5}{3}=\frac{y-(-4)}{7}=\frac{z-6}{2} \]\[\vec a = 5\hat i-4\hat j+6\hat k, \qquad \vec b = 3\hat i+7\hat j+2\hat k \]\[\vec r = \vec a+\lambda\vec b \]

  10. Exercise 40

    The vector equation of the line through the points (3,4,−7)\displaystyle (3,4,-7) and (1,−1,6)\displaystyle (1,-1,6) is ____.

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    NCERT’s answer
    \(\displaystyle (x-3) \hat{i}+(y-4) \hat{j}+(z+7) \hat{k}=\lambda(-2 \hat{i}-5 \hat{j}+13 \hat{k})\)
    \(\displaystyle \vec r = (3\hat i+4\hat j-7\hat k)+\lambda(-2\hat i-5\hat j+13\hat k)\)Position vectors of the two points:\[\vec a = 3\hat i+4\hat j-7\hat k, \qquad \vec b = \hat i-\hat j+6\hat k \]\[\vec b-\vec a = (1-3)\hat i+(-1-4)\hat j+(6+7)\hat k = -2\hat i-5\hat j+13\hat k \]\[\vec r = \vec a+\lambda(\vec b-\vec a) \]