 ॐ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ  या देक्ते हैं कितना दुम हैं। तो तो जाब बात हाती हैं या दि супस तु लींस की तो दो वो वालुम हैं तो बात लेंने off या द्ते गर नी बात आपने चीथ,झो गले बी। तो वालास्ता लाँ with learning and साईन्स, हम को मालुम है, 2 लाईन तीं काटेगरी में गरते है, या तो दोनो पारलल होगे. अगर पारलल हो लें, तो उनका डारेक्ष्झन को साअन्स का जो प्र्पोष्टनालरती कोस्टन्ड होता है, उसे होता है. तो ये जो तर्म होता है, य मद़ा वन लाम्दा तूके साज जुड़े हुए, गो लीन्के देरिक्रूँ को बताते है तो अगर लीन्त परडल हूता, तो ये दोनो वेक्तेस प्रट्टेल होतें बर यहां, दोनो प्रट्टेल नहीं है तो या तो, यी तो नों यों � study ॐ रहे है या तो दोनो ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्योंग ज्य तो हम लोग गा करेंगे पर लिएं दोनो P points को लेलेंगे a1,a2 कहते आंगो तो a1,a2 किया हैं आमारा a1,a2 है kya hai a1,a2 hai icap plus 2j cap plus 3k cap और a2 है 2icap 4j cap plus 5k cap तीकया आम लोग गा करेंगे a1 minus a2 vectr निकालेंगे a2 minus a1 भी आप कर सकते हैं गदाने his.. उब loon ऍंब पर यागटार क भी बी बायाiyaemary , , , , , lii kya plus 5 kya kya toh ju shortest distance hota hai ju shortest distance hota hai shortest distance hum logon hai Patha hi hai is nothing but it's the projection of it's the projection of a1 minus a2 in the direction of in the direction of the cross-product of b1 and b2 chigya toh b1 and b2 keh cross-product mein अगर हम यान मैंवान्स यीटु कर प्रुथेक्षान लेलें। तो वो शेदो basically हमारा शोट्धिस दिस्ट्न्स लेलें। हाता है! औहानाकि मैंसको आपको दिखाने के भी एक कोशिष करोंगा फ्रीट दीजमबत्री के तूं तो यहांपे भी वन क्ऱ बी तो निकालने के लिए हमें क्राऊस पड़क निकालना ख़त्या क्राऊस पडर्क्निकालने का तरीका हम सपको मालुम है हम लोके इक दितरम मनेड बनाते हैन जिस में पहला रो रो ہوता है the unit vectors i cap, j cap, k cap the other row is the components of b1 the third row is your components of b2 let's see what comes out i cap 15 minus 16 minus 1 minus j cap 10 minus 12 minus 2 and plus k cap 8 minus 9 which is minus 1 so this comes out minus i cap plus 2 j cap minus k cap so a1 minus a2 we have got here b1 cross b2 we have got here so let's see what we have to do for the shortest distance projection this is very important projection of a1 minus a2 in the direction of b1 cross b2 so for the shortest distance projection of a1 minus a2 in the direction of b1 cross b2 what does it mean means that we have to take the dot product of a1 minus a2 with the unit vector along b1 cross b2 don't forget this it can be expensive so if you want to take projection then you have to take the dot product unit vector along the along the b1 cross b2 vector okay so let's take it what is the problem in this what was a1 minus a2 I remember to forget i minus 2 j minus 2 k so this was our a1 minus a2 vector b1 cross b2 so it is written above minus i plus 2 j minus k divided by modulus of it modulus of it will be 1 square 2 square 1 square which is actually root 6 so let's see its value will be 1 minus 4 plus 2 okay one more thing here sometimes the answer can be negative so don't forget to take the mod okay mod means to make it positive the modulus of the vector or the magnitude of the mod this is the mod to make it positive so here we will need the mod so it will be 1 by root 6 unit okay so let's see this is our x value so what is the question so what is the question the question is cost inverse cost root 6x okay so let's come cost inverse cost root 6x root 6x means cost inverse cost 1 okay now let's see cost inverse cost theta this is a very important thing which many people do wrong cost inverse cost theta this will be theta only when theta belongs in its principal value branch means between 0 and pi okay so here the 1 radian it will count like 1 radian not like 1 degree so 1 radian of course between 0 and 3.14 okay so it will basically follow our formula so what is cost inverse cost 1 this is the answer so see the answer is coming out so easily let's see which option it is option number C correct answer and you get 5 rupees let's go so this was our mathematically problem solving video now let's see with the help of GeoGebra so friends now we have come on the 3D GeoGebra calculator so here first I will sketch the lines on vector form so basically our line was equal to here in GeoGebra we write capital X and the point which it was passing was 1,2,3 plus lambda times so in GeoGebra we can write any parameter T it is not necessary that you write lambda so 2,3,4 okay so you can see a line on the right side of the screen okay now let's write the other line the other line was our R is equal to 2,4,5 that means the points on it were 2,4,5 plus again T times 3,4,5 3,4,5 okay so you can see both the lines let me show you so these lines are skew lines that means they are non coplanar and they are not available as i told you they are like road or flyover so we have to draw a shorter distance so friends first what i will do here i will try to show the point 1,2,3 so to write the point normally we use our normal trend capital A and 3 coordinates in the bracket you can see here A you can see here okay now let's write the B point which was our B point 2,4,5 let's write it down see this B point is on the other line here what i do let me change the color the color of the first line is a little green yes okay let me show you okay so friends what we have to do we have to draw cross product of the direction vectors of these two lines so what we do for cross we assume two vectors to show the vector in 3d geometry we use small letters so 2,3,4 as soon as i write it it will treat it as a vector although it is not visible and let's write the D vector our D vector is 3,4,5 if it is not visible okay so now we have to show the cross product of COD okay so for cross product there is an order command you will write it and in its parameters and inputs you will write COD vector so you will show the cross product so see cross product dotted line let's make the color a little green let's keep it dark blue okay now friends we have to draw the shortest distance between A and B so first of all make a line connecting A and B okay so you will see that a line segment is made now let me take it come friends now we have positioned the graph that the line that passes from B it will not be visible exactly behind this point B right we have done our view so that particular line passing through the point B is going within the plane of our laptop okay now see what i will do this is the dotted vector because our cross product of COD will translate it means i will move it after moving it i will take it to such a position which will help you in visualization so i will move it to that vector U on point A okay you can see now this distance when you drop from B on this perpendicular line so this distance let me try to show you this distance this distance can you see this distance will be your shortest distance this will be your shortest distance so basically what we do we project the distance between A and B so this length which we have shown this red color which i am trying to show you this is the shadow of A1-A2 so this is the shadow of A1-A2 on the cross product of CND this is the distance which gives us shortest distance so hopefully this video will be helpful for you to visualize how our shortest distance actually looks in 3d geometry thank you so much for watching keep learning and stay safe