Explanation
r=−ˆi+ˆj+ˆk+p(3ˆi−2ˆk)
r=−ˆi+ˆj+ˆk+p(3ˆi+2ˆk)
r=−ˆi+ˆj+ˆk+p(−3ˆi+2ˆk)
r=−ˆi+ˆj+ˆk+p(3ˆi+2ˆj)
Let the plane ax+by+cz+d=0 divides the lines joining (x1,y1,z1) and (x2,y2,z2) in the ratio k:1 as shown in figure.
∴Coordinates of P(kx2+x1k+1,ky2+y1k+1,kz2+z1k+1) must satisfy ax+by+ca+d=0
i.e., (akx2+x1k+1)+b(ky2+y1k+1)+c(kz2+z1k+1)+d=0
⇒a(kx2+x1)+b(ky2+y1)+c(kz2+z1)+c(kz2+z1)+d(k+1)=0
⇒k(ax2+by2+cz2+d)+(ax1+by1+cz1+d)=0
⇒k=−(ax1+by1+cz1+d)(ax2+by2+cz2+d)
Hence, option A is correct.
2x+3y+5z=1 divides (1,0,−3) and (1,−5,7) in the ratio of k:1 at point P.
Then, P(k+1k+1,−5kk+1,7k−3k+1) which must satisfy 2x+3y+5z=1
⇒2(k+1k+1)+3(−5kk+1)+5(7k−3k+1)=1
⇒2k+2−15k+35k−15=k+1
⇒21k=14
⇒k=23
∴2x+3y+5z=1 divides (1,0,−3) and (1,−5,7) in the ratio of 2:3.
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