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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the distance of the point (-1,-5,-10) from the point of intersection of the line \( \overrightarrow r = (2\hat i - \hat j + 2\hat k)+\lambda (3\hat i + 4\hat j + 2\hat k) \) and the plane \( \overrightarrow r = (\hat i - \hat j + \hat k)=5.\)

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  • Distance between two points is $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Step 1:
The equation of the line is $\overrightarrow r=2\hat i-\hat j+2\hat k+\lambda(3\hat i+4\hat j+2\hat k)$-----(1)
The equation of the plane is $\overrightarrow r.(\hat i-\hat j+\hat k)=5$------(2)
On solving equ(1) and (2)
(i.e)$[2\hat i-\hat j+2\hat k+\lambda(3\hat i+4\hat j+2\hat k)].(\hat i-\hat j+\hat k)=5$
On simplifying we get
$(2\hat i-\hat j+2\hat k).(\hat i-\hat j+\hat k)+\lambda(3\hat i+4\hat j+2\hat k).(\hat i-\hat j+\hat k)=5$
Multiplying by applying dot product we get
$(2+1+2)+\lambda(3-4+2)=5$
Step 2:
On simplifying we get
$5+\lambda=5$
Therefore $\lambda=0$
The point of intersection of the line and the plane is $(2\hat i-\hat j+2\hat k)$
The other point is $(-1,-5,-10)=-\hat i-5\hat j-10\hat k$
Step 3:
Distance between two points is $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Substituting for $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ we get,
$\sqrt {(2-(-1))^2+(-1+5)^2+(2-(-10))^2}$
$\Rightarrow \sqrt{3^2+4^2+12^2}$
$\Rightarrow \sqrt{9+16+144}$
$\Rightarrow \sqrt{169}$
$\Rightarrow 13$
Hence the required distance is $13$
answered Nov 11, 2013 by sreemathi.v
 

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