The distance of the point whose position vector is (2 hati+ hatj- hatk) from the plane vecr ⋅( hati-2 hatj+4 hatk)=4 is

Question

The distance of the point whose position vector is (2 hati+ hatj- hatk) from the plane vecr ⋅( hati-2 hatj+4 hatk)=4 is (A) (8/√21) (B) (-8/√21) (C) 8 √21 (D) (-8/21)

Tardigrade Admin · Accepted Answer

Distance =(|2-2-4-4|/√1+4+16)=((8)/√21)

Q. The distance of the point whose position vector is $(2 \hat{i} + \hat{j} - \hat{k})$ from the plane $r \cdot (\hat{i} - 2 \hat{j} + 4 \hat{k}) = 4$ is

$\frac{8}{21}$

$\frac{- 8}{21}$

$821$

$\frac{- 8}{21}$

Distance $= \frac{∣2 - 2 - 4 - 4∣}{1 + 4 + 16} = \frac{( 8 )}{21}$

Q. The distance of the point whose position vector is (2i^+j^​−k^) from the plane r⋅(i^−2j^​+4k^)=4 is

21​8​

21​−8​

821​

21−8​

Distance =1+4+16​∣2−2−4−4∣​=21​(8)​

Q. The distance of the point whose position vector is $(2 \hat{i} + \hat{j} - \hat{k})$ from the plane $r \cdot (\hat{i} - 2 \hat{j} + 4 \hat{k}) = 4$ is

$\frac{8}{21}$

$\frac{- 8}{21}$

$821$

$\frac{- 8}{21}$

Distance $= \frac{∣2 - 2 - 4 - 4∣}{1 + 4 + 16} = \frac{( 8 )}{21}$