Hello friends today we will study problem related to 3d transformation so here is the first problem a homogeneous coordinate point P A point is been given whose coordinate is 3 2 1 A point is will given whose X Y and Z coordinate is given and is translated the x y and z direction so the point which is to be translated along x direction along y direction and along Z direction so translation here it is minus 2 minus 2 and minus 2 means point of each tank point is to be translated minus 2 along X Direction minus 2 units along y direction and minus 2 units along Z direction so first translation transformation is to be done in this problem second respectively followed by successive rotation so rotation is to be done 60 degree about x axis so second transformation which has to be done is rotation 60 degree about x axis next find the final position of homogeneous coordinate so we have to find the final position of homogeneous coordinate so we will start with the solution of this problem so there is a point the point which is to be translated along X direction along y direction and along Z direction and secondly the rotation transformation is to be performed about x axis by 60 degree since it has not been mentioned whether it is to be counter clockwise or clockwise so default we have to consider it as counter clockwise direction so now first transformation is translation so the standard matrices for translation in 3d is given by 1 0 0 0 0 1 0 0 0 0 1 0 T X T y TZ 1 so this is standard transfer translation transformation matrix so for this I will require the value of px and py and PZ so in the problem itself the value of th is managed to the value of T y is minus 2 the value of TZ is also minus 2 so now I will replace the value of T XT Bank TZ the matrices will become 1 0 0 0 0 1 0 0 0 0 1 0 the value of TX is minus 2 value of T Y is minus 2 value of T Z is also minus 2 and this is 1 so now we found the translation matrix now second transformation in this problem is rotation about x axis by 60 degree first translation and second transformation is rotation rotation sixty degree about x-axis so now we will write the standard rotation matrix R which is equal to when 0 0 0 0 cos theta sine theta 0 0 minus sine theta cos theta 0 0 0 0 1 so this is the standard rotation matrices about x-axis cellulitis X R X so now I will require the value of theta so here the value of theta is 60 degree so now I will replace the value of theta s 60 degree so the transformation that is resultant transformation for rotation will become as 1 0 0 0 0 point five point eight six six zero zero minus point is 6 six point five zero zero zero zero one so this is the corresponding matrices for rotation now next we will find the resultant transformation the resultant transformation is the total transformation which is done in this problem so results in transformation which is gained by TR first transformation which is done in this problem is translation and then rotation is is been done so TR which is equal to I will write the matrices for translation which is 1 0 0 0 0 1 0 0 0 0 1 0 minus 2 minus 2 minus 2 1 and then the rotation transformation matrices which is 1 0 0 0 0 point 5 point 8 6 6 0 0 minus point eight six six point five zero zero zero zero five so now I will multiply both this matrices then the final transformation that is resultant transformation which I will get as 1 0 0 0 0 point five point eight six six zero zero minus point eight six six point five zero minus two point seven three two minus two point seven three two one so this is the resident transformation matrix in this problem we have to find the final position so we have to find the final position so we know the formula for final position as final position which is equal to initial position into TR so we know the initial position of P which is 3 comma 2 comma 1 so now I will convert the coordinate form of P into matrices form then it will become 3 2 1 1 now I will multiply it with P R so the matrices for TR is 1 0 0 0 0 point five point eight six six zero zero minus point eight six six point five zero minus two point seven three two one seven three two one so now so now I will multiply the both the matrices so that I will get final position that is P – I am representing – for final position so after multiplying both the matrices the value which I get is 1 0.864 minus 0.5 1 so this is a final position now I will convert this final position which is of the form that is in the matrices form so I will convert this matrices form into coordinate form so I’ll write as final position P – that is converting into coordinate form which is 1 comma 0.864 comma minus 0.5 so this is solution for the problem thank you

very helpful video sir

Hei can u explain me how cos(60) value in rotation matrix got the value 0.5 and sin(60) got 0.866 value

Bhai ju samjhana wah Kab samhu ge p

Remove private videos plz soon exam is on

Sir, here you have taken both the matrices (for translation and rotation) as horizontal matrices. And then you have taken the original coordinates also in a horizontal or row matrix.If I am taking it in column matrix and applying all the other matrices of rotation and translation also in the same manner, then I am not getting the same transformation matrix. Why ?

I feel doubt in that rotation matrix in counterclockwise..direction sir..the sine element of the second column should have the negative sign rather than positive…

Very helpful . Thanx sir.

sir, last mai ads kyu daal dete ho problem hoti padne mai

Sir intial position [3 2 1 1] kaise find kiya plz tell mi

Bakwas, Sirf chamak dhamak

Your channel name in youtube

0.866 not 0.864

Main problem to bta hi nhi rhe hi

Cadê os br aí? hahaha