a, b, c are the pth, qth, rth terms of an H.P. and u= (q-r) hati+(r-p) hatj+(p-q) hatk v= ( hati/a)+ ( hatj/b)+ ( hatk/c), then

Question

a, b, c are the pth, qth, rth terms of an H.P. and u= (q-r) hati+(r-p) hatj+(p-q) hatk v= ( hati/a)+ ( hatj/b)+ ( hatk/c), then (A) u,v are parallel vectors (B) u,v are orthogonal vectors (C) u.v=1 (D) u×v= hati + hatj + hatk

Tardigrade Admin · Accepted Answer

vecu . vecv=(q-r/a)+(r-p/b)+(p-q/c) Since (1/a)=x+(p-1)y (1/b)=x+(q-1)y (1/c)=x+(r-1)y Putting in (1), we get vecu . vecv=(q-r)[x+(p-1)y]+(r-p) [x+(q-1)y]+(p-q)[x+(r-1)y]=0 ∴ vecu . vecv=0 ldots(1) ∴ vecu . vecv are orthogonal vectors

Q. a, b, c are the pth, qth, rth terms of an H.P. and $u = (q - r) \hat{i} + (r - p) \hat{j} + (p - q) \hat{k} v = \frac{i ^}{a} + \frac{j ^}{b} + \frac{k ^}{c},$ then

$u, v$ are parallel vectors

$u, v$ are orthogonal vectors

$u . v = 1$

$u \times v = \hat{i} + \hat{j} + \hat{k}$

Q. a, b, c are the pth, qth, rth terms of an H.P. and u=(q−r)i^+(r−p)j^​+(p−q)k^v=ai^​+bj^​​+ck^​, then

u,v are parallel vectors

u,v are orthogonal vectors

u.v=1

u×v=i^+j^​+k^

u.v=aq−r​+br−p​+cp−q​ Since a1​=x+(p−1)y b1​=x+(q−1)y c1​=x+(r−1)y Putting in (1), we get u.v=(q−r)[x+(p−1)y]+(r−p) [x+(q−1)y]+(p−q)[x+(r−1)y]=0 ∴u.v=0…(1) ∴u.v are orthogonal vectors

Q. a, b, c are the pth, qth, rth terms of an H.P. and $u = (q - r) \hat{i} + (r - p) \hat{j} + (p - q) \hat{k} v = \frac{i ^}{a} + \frac{j ^}{b} + \frac{k ^}{c},$ then

$u, v$ are parallel vectors

$u, v$ are orthogonal vectors

$u . v = 1$

$u \times v = \hat{i} + \hat{j} + \hat{k}$