QuestionsQuestion 1Compute the cross product a×b\mathbf{a} \times \mathbf{b}a×b and b×a\mathbf{b} \times \mathbf{a}b×a. Check your answer by showing that each a\mathbf{a}a and b\mathbf{b}b are perpendicular to a×b\mathbf{a} \times \mathbf{b}a×b and b×a\mathbf{b} \times \mathbf{a}b×a.(i)a=2i+j−k\mathbf{a} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}a=2i+j−k, b=i−j+k\mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k}b=i−j+k(ii)a=i+3j+2k\mathbf{a} = \mathbf{i} + 3\mathbf{j} + 2\mathbf{k}a=i+3j+2k, b=2i−j+k\mathbf{b} = 2\mathbf{i} - \mathbf{j} + \mathbf{k}b=2i−j+k(iii)a=2i−2j+k\mathbf{a} = 2\mathbf{i} - 2\mathbf{j} + \mathbf{k}a=2i−2j+k, b=−i+j+3k\mathbf{b} = -\mathbf{i} + \mathbf{j} + 3\mathbf{k}b=−i+j+3k(iv)a=−4i+j−2k\mathbf{a} = -4\mathbf{i} + \mathbf{j} - 2\mathbf{k}a=−4i+j−2k, b=2i+j+k\mathbf{b} = 2\mathbf{i} + \mathbf{j} + \mathbf{k}b=2i+j+kSolutionTheoryQuestion 2Find a unit vector perpendicular to the plane containing a\mathbf{a}a and b\mathbf{b}b. Also find sine of the angle between them:(i)a=i+6j−3k\mathbf{a} = \mathbf{i} + 6\mathbf{j} - 3\mathbf{k}a=i+6j−3k, b=2i+j+3k\mathbf{b} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k}b=2i+j+3k(ii)a=−i−j−k\mathbf{a} = -\mathbf{i} - \mathbf{j} - \mathbf{k}a=−i−j−k, b=2i−3j+4k\mathbf{b} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}b=2i−3j+4k(iii)a=i+j+k\mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k}a=i+j+k, b=i−j−k\mathbf{b} = \mathbf{i} - \mathbf{j} - \mathbf{k}b=i−j−k(iv)a=5i+j−3k\mathbf{a} = 5\mathbf{i} + \mathbf{j} - 3\mathbf{k}a=5i+j−3k, b=−2i+4j+k\mathbf{b} = -2\mathbf{i} + 4\mathbf{j} + \mathbf{k}b=−2i+4j+kSolutionTheoryQuestion 3Find the area of the triangle, formed by the points P,QP, QP,Q and RRR.(i)P(2,3,5);Q(1,2,0);R(4,1,2)P(2, 3, 5); Q(1, 2, 0); R(4, 1, 2)P(2,3,5);Q(1,2,0);R(4,1,2)(ii)P(0,0,1);Q(2,−1,2);R(−1,3,2)P(0, 0, 1); Q(2, -1, 2); R(-1, 3, 2)P(0,0,1);Q(2,−1,2);R(−1,3,2)SolutionTheoryQuestion 4Find the area of a parallelogram, whose vertices are:(i)A(1,1,1);B(4,2,3);C(5,6,7);D(2,5,5)A(1, 1, 1); B(4, 2, 3); C(5, 6, 7); D(2, 5, 5)A(1,1,1);B(4,2,3);C(5,6,7);D(2,5,5)(ii)A(4,5,6);B(1,3,2);C(−2,0,1);D(1,2,5)A(4, 5, 6); B(1, 3, 2); C(-2, 0, 1); D(1, 2, 5)A(4,5,6);B(1,3,2);C(−2,0,1);D(1,2,5)SolutionTheoryQuestion 5If the cross product of the vectors u=7i−4j+5k\mathbf{u} = 7\mathbf{i} - 4\mathbf{j} + 5\mathbf{k}u=7i−4j+5k and v=ai−bj+3k\mathbf{v} = a\mathbf{i} - b\mathbf{j} + 3\mathbf{k}v=ai−bj+3k is zero, then find the values of aaa and bbb.SolutionTheoryQuestion 6Which vectors, if any, are perpendicular or parallel(i)u=5i−j+k;v=j−5k;w=−15i+3j−3k\mathbf{u} = 5\mathbf{i} - \mathbf{j} + \mathbf{k}; \mathbf{v} = \mathbf{j} - 5\mathbf{k}; \mathbf{w} = -15\mathbf{i} + 3\mathbf{j} - 3\mathbf{k}u=5i−j+k;v=j−5k;w=−15i+3j−3k(ii)u=i+2j−k;v=−i+j+k;w=−π2i−πj+π2k\mathbf{u} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}; \mathbf{v} = -\mathbf{i} + \mathbf{j} + \mathbf{k}; \mathbf{w} = -\dfrac{\pi}{2}\mathbf{i} - \pi\mathbf{j} + \dfrac{\pi}{2}\mathbf{k}u=i+2j−k;v=−i+j+k;w=−2πi−πj+2πkSolutionTheoryQuestion 7Use the definition of cross product, for any vectors u,v,w\mathbf{u}, \mathbf{v}, \mathbf{w}u,v,w and scalar kkk, prove that(i)u×(−u)=0\mathbf{u} \times (-\mathbf{u}) = 0u×(−u)=0(ii)u×v=−v×u\mathbf{u} \times \mathbf{v} = -\mathbf{v} \times \mathbf{u}u×v=−v×u(iii)u×(kv)=(ku)×v=k(u×v)\mathbf{u} \times (k\mathbf{v}) = (k\mathbf{u}) \times \mathbf{v} = k(\mathbf{u} \times \mathbf{v})u×(kv)=(ku)×v=k(u×v)(iv)u×(v+w)=(u×v)+(u×w)\mathbf{u} \times (\mathbf{v} + \mathbf{w}) = (\mathbf{u} \times \mathbf{v}) + (\mathbf{u} \times \mathbf{w})u×(v+w)=(u×v)+(u×w)SolutionTheoryQuestion 8Prove that: a×(b+c)+b×(c+a)+c×(a+b)=0\mathbf{a} \times (\mathbf{b} + \mathbf{c}) + \mathbf{b} \times (\mathbf{c} + \mathbf{a}) + \mathbf{c} \times (\mathbf{a} + \mathbf{b}) = 0a×(b+c)+b×(c+a)+c×(a+b)=0.SolutionTheoryQuestion 9If a+b+c=0\mathbf{a} + \mathbf{b} + \mathbf{c} = 0a+b+c=0, then prove that a×b=b×c=c×a\mathbf{a} \times \mathbf{b} = \mathbf{b} \times \mathbf{c} = \mathbf{c} \times \mathbf{a}a×b=b×c=c×a.SolutionTheoryQuestion 10Prove that: sin(α−β)=sinαcosβ+cosαsinβ\sin(\alpha - \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\betasin(α−β)=sinαcosβ+cosαsinβ.SolutionTheoryQuestion 11Show that ∣a×b∣2=∣a∣2∣b∣2−(a⋅b)2|\mathbf{a} \times \mathbf{b}|^2 = |\mathbf{a}|^2 |\mathbf{b}|^2 - (\mathbf{a} \cdot \mathbf{b})^2∣a×b∣2=∣a∣2∣b∣2−(a⋅b)2.SolutionTheoryQuestion 12Use the definition of cross product, prove that for any vectors u\mathbf{u}u and v\mathbf{v}v (u+v)×(u−v)=−2(u×v)(\mathbf{u} + \mathbf{v}) \times (\mathbf{u} - \mathbf{v}) = -2(\mathbf{u} \times \mathbf{v})(u+v)×(u−v)=−2(u×v).SolutionTheoryQuestion 13Find the moment about the point M(1,−3,3)M(1, -3, 3)M(1,−3,3) of the force represented by AB→\overrightarrow{AB}AB, where the coordinates of points A(4,3,−1)A(4, 3, -1)A(4,3,−1) and B(−1,3,7)B(-1, 3, 7)B(−1,3,7) are given.SolutionTheoryQuestion 14A force F=6i+4j−4k\mathbf{F} = 6\mathbf{i} + 4\mathbf{j} - 4\mathbf{k}F=6i+4j−4k is applied at the point A(1,−1,2)A(1, -1, 2)A(1,−1,2). Find the moment of the force about the point B(3,−2,3)B(3, -2, 3)B(3,−2,3).SolutionTheoryQuestion 15Give a force F=2i+j−3k\mathbf{F} = 2\mathbf{i} + \mathbf{j} - 3\mathbf{k}F=2i+j−3k acting at a point A(1,−2,1)A(1, -2, 1)A(1,−2,1). Find the moment of F\mathbf{F}F about the point B(2,0,−2)B(2, 0, -2)B(2,0,−2).SolutionTheoryQuestion 16A force F=−2i+j−3k\mathbf{F} = -2\mathbf{i} + \mathbf{j} - 3\mathbf{k}F=−2i+j−3k is applied at P(−1,−3,2)P(-1, -3, 2)P(−1,−3,2). Find its moment about the point Q(4,2,2)Q(4, 2, 2)Q(4,2,2).SolutionTheory