Skip to content
Question of 72

Q.(i) [1] Give an example of any one point which lie in second octant.

(ii) [3] Show that the points A(0,7,10)A(0, 7, 10), B(−1,6,6)B(-1, 6, 6) and C(−4,9,6)C(-4, 9, 6) are the vertices of a right angled triangle.
Kerala DhseKerala DHSE Plus One Board 2024Subjective· 4mImportance★★★★★
0% · 0/72 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

(i) The second octant has x<0,y>0,z>0x<0, y>0, z>0. (ii) Use the distance formula on all three sides and check whether the Pythagoras relation holds.

(i) The octants are determined by the signs of x,y,zx,y,z. In the second octant, x<0x<0, y>0y>0, z>0z>0.

An example point: (−1,2,3)(-1,2,3).

(ii) Given A(0,7,10)A(0,7,10), B(−1,6,6)B(-1,6,6), C(−4,9,6)C(-4,9,6). Distance formula: d=(x2−x1)2+(y2−y1)2+(z2−z1)2d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}.

AB2=(0−(−1))2+(7−6)2+(10−6)2=1+1+16=18.AB^2 = (0-(-1))^2+(7-6)^2+(10-6)^2 = 1+1+16 = 18.

BC2=(−1−(−4))2+(6−9)2+(6−6)2=9+9+0=18.BC^2 = (-1-(-4))^2+(6-9)^2+(6-6)^2 = 9+9+0 = 18.

CA2=(−4−0)2+(9−7)2+(6−10)2=16+4+16=36.CA^2 = (-4-0)^2+(9-7)^2+(6-10)^2 = 16+4+16 = 36.

Check: AB2+BC2=18+18=36=CA2AB^2+BC^2 = 18+18 = 36 = CA^2.

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.