 Good morning friends, I am Purva and today we will discuss the following question. Let vector A is equal to i cap plus 4 j cap plus 2 k cap, vector B is equal to 3 i cap minus 2 j cap plus 7 k cap and vector C is equal to 2 i cap minus j cap plus 4 k cap. Find a vector vector D which is perpendicular to both vector A and vector B and vector C dot vector D is equal to 15. Let us now begin with the solution. Now let vector D is equal to D 1 i cap plus D 2 j cap plus D 3 k cap B perpendicular to vector A and vector B. Now since vector D is perpendicular to vector A, therefore we have vector D dot vector A is equal to 0. This implies vector D is D 1 i cap plus D 2 j cap plus D 3 k cap dot vector A is i cap plus 4 j cap plus 2 k cap is equal to 0. This implies D 1 into 1 is D 1 plus D 2 into 4 is 4 D 2 plus D 3 into 2 is 2 D 3 is equal to 0. We mark this as equation 1. Now again since vector D is perpendicular to vector B, therefore we have vector D dot vector B is equal to 0. This implies now vector D is D 1 i cap plus D 2 j cap plus D 3 k cap dot vector B is 3 i cap minus 2 j cap plus 7 k cap and this is equal to 0. This implies D 1 into 3 is 3 D 1, D 2 into minus 2 is minus 2 D 2 and D 3 into 7 gives plus 7 D 3 is equal to 0. We mark this as equation 2. We are also given that vector C dot vector D is equal to 15. Now vector C is equal to 2 i cap minus j cap plus 4 k cap dot vector D is D 1 i cap plus D 2 j cap plus D 3 k cap is equal to 15. Now this implies 2 into D 1 gives 2 D 1 minus 1 into D 2 gives minus D 2 and 4 into D 3 gives plus 4 D 3 is equal to 15. We mark this as equation 3. Now solving equation 1 and 2 using cross-multiplication method we get D 1 upon 28 plus 4 is equal to D 2 upon 6 minus 7 is equal to D 3 upon minus 2 minus 12 and so we get D 1 upon 32 is equal to D 2 upon minus 1 which is equal to D 3 upon minus 14 and this is equal to say some lambda and this implies D 1 is equal to 32 lambda D 2 is equal to minus lambda and D 3 is equal to minus 14 lambda now substituting these values of D 1 D 2 and D 3 in equation 3 we get 2 into 32 lambda plus lambda plus 4 into minus 14 lambda is equal to 15 this implies 64 lambda plus lambda minus 56 lambda is equal to 15 this further implies 9 lambda is equal to 15 which implies lambda is equal to 15 upon 9 and we get this implies lambda is equal to 5 upon 3 now D 1 is equal to 32 lambda so we get this implies D 1 is equal to 32 into 5 by 3 since lambda is equal to 5 by 3 and this implies D 1 is equal to 160 upon 3 D 2 is equal to minus lambda so we get this implies D 2 is equal to minus 5 upon 3 and D 3 is equal to minus 14 lambda so we get this implies D 3 is equal to minus 14 into 5 by 3 which implies D 3 is equal to minus 70 upon 3 therefore the required vector D is equal to 160 upon 3 i cap minus 5 upon 3 j cap minus 70 upon 3 k cap hence we write our answer as 1 upon 3 into 160 i cap minus 5 j cap plus 70 k cap hope you have understood the solution bye and take care