A plane P passes through the point (1,1 , 1) and is parallel to the vectors overset arrow a=- hati+ hatj and overset arrow b= hati- hatk . The distance of the point ((3 √3/2) , 3 √3 , 3) from the plane is equal to

Question

A plane P passes through the point (1,1 , 1) and is parallel to the vectors overset arrow a=- hati+ hatj and overset arrow b= hati- hatk . The distance of the point ((3 √3/2) , 3 √3 , 3) from the plane is equal to (A) √3 units (B) (9/2) units (C) 3√3 units (D) 3 units

Tardigrade Admin · Accepted Answer

Equation of the plane is | x-1 y-1 z-1 -1 1 0 1 0 -1 |=0 ⇒ (x - 1)(- 1)-(y - 1)(1)+(z - 1)(- 1)=0 ⇒ x+y+z-3=0 The distance of point ((3 √3/2) , 3 √3 , 3) from x+y+z-3=0 is |((3 √3/2) + 3 √3 + 3 - 3/√3)|=(9/2) units

Q. A plane $P$ passes through the point $(1, 1, 1)$ and is parallel to the vectors $a \to = - \hat{i} + \hat{j}$ and $b \to = \hat{i} - \hat{k}$ . The distance of the point $(\frac{3 3}{2}, 33, 3)$ from the plane is equal to

$3$ units

$\frac{9}{2}$ units

$33$ units

$3$ units

Q. A plane P passes through the point (1,1,1) and is parallel to the vectors a→=−i^+j^​ and b→=i^−k^ . The distance of the point (233​​,33​,3) from the plane is equal to

3​ units

29​ units

33​ units

3 units

Equation of the plane is ∣∣​x−1−11​y−110​z−10−1​∣∣​=0 ⇒(x−1)(−1)−(y−1)(1)+(z−1)(−1)=0 ⇒x+y+z−3=0 The distance of point (233​​,33​,3) from x+y+z−3=0 is ∣∣​3​233​​+33​+3−3​∣∣​=29​ units

Q. A plane $P$ passes through the point $(1, 1, 1)$ and is parallel to the vectors $a \to = - \hat{i} + \hat{j}$ and $b \to = \hat{i} - \hat{k}$ . The distance of the point $(\frac{3 3}{2}, 33, 3)$ from the plane is equal to

$3$ units

$\frac{9}{2}$ units

$33$ units

$3$ units