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NCERT Exemplar · Class 12 Mathematics Three Dimensional Geometry

49 questions · 49 still being checked

EXERCISE 11.3 41–49 (part 5 of 5)

  1. Fill in the blanks in each of the Exercises $\displaystyle 37$ to 41.

    Exercise 41

    The cartesian equation of the plane r⃗⋅(i^+j^−k^)=2\displaystyle \vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=2 is ____.

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    NCERT’s answer
    \(\displaystyle x+y-z=2\)
    \(\displaystyle x+y-z=2\)Put \(\displaystyle \vec r = x\hat i+y\hat j+z\hat k\).\[(x\hat i+y\hat j+z\hat k)\cdot(\hat i+\hat j-\hat k)=2 \]\[x+y-z=2 \]
  2. State True or False for the statements in each of the Exercises $\displaystyle 42$ to 49.

    Exercise 42

    The unit vector normal to the plane x+2y+3z−6=0\displaystyle x+2 y+3 z-6=0 is 114i^+214j^+314k^\displaystyle \frac{1}{\sqrt{14}} \hat{i}+\frac{2}{\sqrt{14}} \hat{j}+\frac{3}{\sqrt{14}} \hat{k}.

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    NCERT’s answer
    True
    True. The coefficients of \(\displaystyle x,y,z\) give a normal; dividing by its length gives a unit normal.\[\vec n = \hat i+2\hat j+3\hat k, \qquad |\vec n| = \sqrt{1+4+9} = \sqrt{14} \]\[\hat n = \frac{\vec n}{|\vec n|} = \frac{1}{\sqrt{14}}\hat i+\frac{2}{\sqrt{14}}\hat j+\frac{3}{\sqrt{14}}\hat k \]\[x+2y+3z=6 \;\Rightarrow\; \vec r\cdot\hat n=\frac{6}{\sqrt{14}} \]

  3. Exercise 43

    The intercepts made by the plane 2x−3y+5z+4=0\displaystyle 2 x-3 y+5 z+4=0 on the co-ordinate axis are −2,43,−45\displaystyle -2, \frac{4}{3},-\frac{4}{5}.

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    NCERT’s answer
    True
    True — set two of \(\displaystyle x,y,z\) to zero in turn.\[2x-3y+5z+4=0 \Rightarrow 2x-3y+5z=-4 \] \[y=z=0:\quad 2x=-4 \Rightarrow x=-2 \] \[x=z=0:\quad -3y=-4 \Rightarrow y=\frac{4}{3} \] \[x=y=0:\quad 5z=-4 \Rightarrow z=-\frac{4}{5} \]

  4. Exercise 44

    The angle between the line r⃗=(5i^−j^−4k^)+λ(2i^−j^+k^)\displaystyle \vec{r}=(5 \hat{i}-\hat{j}-4 \hat{k})+\lambda(2 \hat{i}-\hat{j}+\hat{k}) and the plane r⃗.(3i^−4j^−k^)+5=0\displaystyle \vec{r} .(3 \hat{i}-4 \hat{j}-\hat{k})+5=0 is sin⁡−1(5291)\displaystyle \sin ^{-1}\left(\frac{5}{2 \sqrt{91}}\right).

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    NCERT’s answer
    False
    False — \(\displaystyle \sin\theta=\dfrac{|\vec b\cdot\vec n|}{|\vec b|\,|\vec n|}\) with \(\displaystyle \vec b\) the line's direction and \(\displaystyle \vec n\) the plane's normal.\[\vec b=2\hat i-\hat j+\hat k,\qquad \vec n=3\hat i-4\hat j-\hat k \] \[\vec b\cdot\vec n=6+4-1=9 \] \[|\vec b|=\sqrt{6},\qquad |\vec n|=\sqrt{9+16+1}=\sqrt{26} \] \[\sin\theta=\frac{9}{\sqrt{6}\,\sqrt{26}}=\frac{9}{2\sqrt{39}} \] \[\frac{9}{2\sqrt{39}}\neq\frac{5}{2\sqrt{91}} \]

  5. Exercise 45

    The angle between the planes r⃗.(2i^−3j^+k^)=1\displaystyle \vec{r} .(2 \hat{i}-3 \hat{j}+\hat{k})=1 and rˉ.(i^−j^)=4\displaystyle \bar{r} .(\hat{i}-\hat{j})=4 is cos⁡−1(−558)\displaystyle \cos ^{-1}\left(\frac{-5}{\sqrt{58}}\right).

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    NCERT’s answer
    False
    False — the angle between the planes is the angle between their normals.\[\vec n_1=2\hat i-3\hat j+\hat k,\qquad \vec n_2=\hat i-\hat j \] \[\vec n_1\cdot\vec n_2=2+3+0=5 \] \[|\vec n_1|=\sqrt{14},\qquad |\vec n_2|=\sqrt{2} \] \[\cos\theta=\frac{5}{\sqrt{14}\,\sqrt{2}}=\frac{5}{2\sqrt{7}} \] \[\frac{5}{2\sqrt{7}}\neq\frac{-5}{\sqrt{58}} \]

  6. Exercise 46

    The line r⃗=2i^−3j^−k^+λ(i^−j^+2k^)\displaystyle \vec{r}=2 \hat{i}-3 \hat{j}-\hat{k}+\lambda(\hat{i}-\hat{j}+2 \hat{k}) lies in the plane r⃗.(3i^+j^−k^)+2=0\displaystyle \vec{r} .(3 \hat{i}+\hat{j}-\hat{k})+2=0.

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    False — the line is parallel to the plane, but its point is not on the plane.\[\vec b=\hat i-\hat j+2\hat k,\qquad \vec n=3\hat i+\hat j-\hat k \] \[\vec b\cdot\vec n=3-1-2=0 \]The plane is \(\displaystyle 3x+y-z+2=0\); test the line's point \(\displaystyle (2,-3,-1)\):\[3(2)+(-3)-(-1)+2=6\neq 0 \]

  7. Exercise 47

    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 r⃗=5i^−4j^+6k^+λ(3i^+7j^+2k^)\displaystyle \vec{r}=5 \hat{i}-4 \hat{j}+6 \hat{k}+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k}).

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    NCERT’s answer
    True
    True — the line passes through \(\displaystyle (5,-4,6)\) with direction ratios \(\displaystyle 3,7,2\).\[\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}=\lambda \] \[x=5+3\lambda,\quad y=-4+7\lambda,\quad z=6+2\lambda \] \[\vec r=5\hat i-4\hat j+6\hat k+\lambda(3\hat i+7\hat j+2\hat k) \]

  8. Exercise 48

    The equation of a line, which is parallel to 2i^+j^+3k^\displaystyle 2 \hat{i}+\hat{j}+3 \hat{k} and which passes through the point (5\displaystyle 5,-2,4\displaystyle 2{,}4), is x−52=y+2−1=z−43\displaystyle \frac{x-5}{2}=\frac{y+2}{-1}=\frac{z-4}{3}.

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    NCERT’s answer
    False
    False — the direction ratios must be \(\displaystyle 2,1,3\), but the given line has \(\displaystyle 2,-1,3\).\[\frac{x-5}{2}=\frac{y+2}{1}=\frac{z-4}{3} \] \[\frac{2}{2}=1\neq\frac{-1}{1}=-1 \]

  9. Exercise 49

    If the foot of perpendicular drawn from the origin to a plane is (5,−3,−2)\displaystyle (5,-3,-2), then the equation of plane is r⃗⋅(5i^−3j^−2k^)=38\displaystyle \vec{r} \cdot(5 \hat{i}-3 \hat{j}-2 \hat{k})=38.

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    NCERT’s answer
    True
    True — the normal is the vector from the origin to the foot \(\displaystyle (5,-3,-2)\), and the plane passes through the foot.\[\vec n=5\hat i-3\hat j-2\hat k,\qquad \vec a=5\hat i-3\hat j-2\hat k \] \[\vec r\cdot\vec n=\vec a\cdot\vec n \] \[\vec a\cdot\vec n=25+9+4=38 \] \[\vec r\cdot(5\hat i-3\hat j-2\hat k)=38 \]