12ஆம் வகுப்பு கணிதவியல் — வெக்டர் இயற்கணிதத்தின் பயன்பாடுகள்: Online Practice Test

Explanation. Parallel vectors satisfy \(\vec{a}\times\vec{b}=\vec{0}\). Interchanging two entries of a scalar triple product only flips its sign, so \([\vec{a},\vec{c},\vec{b}]=-[\vec{a},\vec{b},\vec{c}]=-(\vec{a}\times\vec{b})\cdot\vec{c}\). Because \(\vec{a}\times\vec{b}=\vec{0}\), this equals \(0\). Equivalently, three vectors of which two are parallel are coplanar, and the box product of coplanar vectors vanishes.

Explanation. Any vector lying in the plane of \(\vec{\beta}\) and \(\vec{\gamma}\) can be written as \(\vec{\alpha}=s\vec{\beta}+t\vec{\gamma}\). The three vectors are then coplanar, and the scalar triple product of coplanar vectors is always zero, so \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=0\). Geometrically the box product is the volume of the parallelepiped on the three vectors, which collapses to zero when they share a plane.

Explanation. The three conditions mean the vectors are mutually perpendicular. For an orthogonal triad, \(\vec{b}\times\vec{c}\) is aligned with \(\vec{a}\) and has magnitude \(|\vec{b}|\,|\vec{c}|\). Hence \(\big|[\vec{a},\vec{b},\vec{c}]\big|=\big|\vec{a}\cdot(\vec{b}\times\vec{c})\big|=|\vec{a}|\,|\vec{b}|\,|\vec{c}|\), the volume of the rectangular box with the three vectors as edges.

Explanation. Apply the vector triple product identity \(\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}\). Since \(\vec{a}\parallel\vec{c}\) and both are unit vectors, \(\vec{a}\cdot\vec{c}=1\); since \(\vec{a}\perp\vec{b}\), \(\vec{a}\cdot\vec{b}=0\). The expression collapses to \(1\cdot\vec{b}-0\cdot\vec{c}=\vec{b}\).

Explanation. Each numerator is a cyclic rearrangement of \([\vec{a},\vec{b},\vec{c}]\), so each equals \(1\). Among the denominators, \((\vec{c}\times\vec{a})\cdot\vec{b}=[\vec{c},\vec{a},\vec{b}]=1\) and \((\vec{a}\times\vec{b})\cdot\vec{c}=[\vec{a},\vec{b},\vec{c}]=1\) are cyclic permutations, but \((\vec{c}\times\vec{b})\cdot\vec{a}=[\vec{c},\vec{b},\vec{a}]=-1\) is an odd permutation. The sum is therefore \(\tfrac{1}{1}+\tfrac{1}{1}+\tfrac{1}{-1}=1\).

Explanation. The volume is the magnitude of the determinant whose rows are the three edge vectors: \[\begin{vmatrix}1&1&0\\1&2&0\\1&1&\pi\end{vmatrix}.\] Expanding along the third column, only the entry \(\pi\) contributes, giving \(\pi\,(1\cdot2-1\cdot1)=\pi\). The volume is \(\pi\) cubic units.

Explanation. Here \([\vec{a},\vec{b},\vec{a}\times\vec{b}]=(\vec{a}\times\vec{b})\cdot(\vec{a}\times\vec{b})=|\vec{a}\times\vec{b}|^{2}\). For unit vectors \(|\vec{a}\times\vec{b}|=\sin\theta\), so the box product equals \(\sin^{2}\theta\). Setting \(\sin^{2}\theta=\tfrac14\) gives \(\sin\theta=\tfrac12\), hence \(\theta=\tfrac{\pi}{6}\).

Explanation. First \(\vec{a}\times\vec{b}=-\hat{i}+\hat{j}\). Crossing with \(\vec{c}=\hat{i}\) gives \((-\hat{i}+\hat{j})\times\hat{i}=-\hat{k}\). Writing \(\lambda\vec{a}+\mu\vec{b}=(\lambda+\mu)\hat{i}+(\lambda+\mu)\hat{j}+\lambda\hat{k}\) and matching with \(-\hat{k}\): the \(\hat{k}\) term gives \(\lambda=-1\) and the \(\hat{i}\) term gives \(\lambda+\mu=0\), so \(\mu=1\). Therefore \(\lambda+\mu=0\).

Explanation. A standard identity gives \([\vec{a}\times\vec{b},\ \vec{b}\times\vec{c},\ \vec{c}\times\vec{a}]=[\vec{a},\vec{b},\vec{c}]^{2}\). With \([\vec{a},\vec{b},\vec{c}]=3\), the inner box product equals \(3^{2}=9\). Squaring as the question asks gives \(9^{2}=81\).

Explanation. Expanding the left side, \(\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}\). Comparing with \(\dfrac{\vec{b}+\vec{c}}{\sqrt{2}}\) and using that \(\vec{b},\vec{c}\) are independent, \(\vec{a}\cdot\vec{c}=\tfrac{1}{\sqrt2}\) and \(\vec{a}\cdot\vec{b}=-\tfrac{1}{\sqrt2}\). Since all are unit vectors, \(\cos\theta=\vec{a}\cdot\vec{b}=-\tfrac{1}{\sqrt2}\), giving \(\theta=\tfrac{3\pi}{4}\).

Explanation. Write \(V=[\vec{a},\vec{b},\vec{c}]\). The first volume satisfies \([\vec{a}\times\vec{b},\vec{b}\times\vec{c},\vec{c}\times\vec{a}]=V^{2}=8\). For the second set, the identity \([(\vec{a}\times\vec{b})\times(\vec{b}\times\vec{c}),\ (\vec{b}\times\vec{c})\times(\vec{c}\times\vec{a}),\ (\vec{c}\times\vec{a})\times(\vec{a}\times\vec{b})]=[\vec{a}\times\vec{b},\vec{b}\times\vec{c},\vec{c}\times\vec{a}]^{2}=V^{4}\) applies. Hence the volume is \((V^{2})^{2}=8^{2}=64\) cubic units.

Explanation. The vector \(\vec{a}\times\vec{b}\) is normal to \(P_{1}\) and \(\vec{c}\times\vec{d}\) is normal to \(P_{2}\). The condition \((\vec{a}\times\vec{b})\times(\vec{c}\times\vec{d})=\vec{0}\) means these two normals are parallel. Planes whose normals are parallel are themselves parallel, so the angle between them is \(0\).

Explanation. Expand both sides. The left gives \((\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}\); the right gives \((\vec{a}\cdot\vec{c})\vec{b}-(\vec{b}\cdot\vec{c})\vec{a}\). Cancelling the common term \((\vec{a}\cdot\vec{c})\vec{b}\) leaves \((\vec{a}\cdot\vec{b})\vec{c}=(\vec{b}\cdot\vec{c})\vec{a}\). Because \(\vec{a}\cdot\vec{b}\neq0\), this forces \(\vec{c}=\dfrac{\vec{b}\cdot\vec{c}}{\vec{a}\cdot\vec{b}}\,\vec{a}\), a nonzero scalar multiple of \(\vec{a}\). Thus \(\vec{a}\) and \(\vec{c}\) are parallel.

Explanation. A vector perpendicular to \(\vec{a}\) but lying in the plane of \(\vec{b}\) and \(\vec{c}\) is \(\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}\). Here \(\vec{a}\cdot\vec{c}=6+15+1=22\) and \(\vec{a}\cdot\vec{b}=2+6+5=13\), so the vector is \(22\vec{b}-13\vec{c}\). Componentwise, \(22(\hat{i}+2\hat{j}-5\hat{k})-13(3\hat{i}+5\hat{j}-\hat{k})=-17\hat{i}-21\hat{j}-97\hat{k}\).

Explanation. The first line has direction ratios \((3,-2,0)\) because \(z=2\) is constant. For the second, rewriting \(\dfrac{2y+3}{3}=\dfrac{y+3/2}{3/2}\) gives direction ratios proportional to \((1,\tfrac32,2)\), i.e. \((2,3,4)\). Their dot product is \(3\cdot2+(-2)\cdot3+0\cdot4=0\), so the lines are perpendicular and the angle is \(\tfrac{\pi}{2}\).

Explanation. For the line to lie in the plane, its direction \((3,-5,2)\) must be perpendicular to the plane's normal \((1,3,-\alpha)\): \(3-15-2\alpha=0\Rightarrow\alpha=-6\). The point \((2,1,-2)\) on the line must also satisfy the plane: \(2+3(1)-\alpha(-2)+\beta=0\). Substituting \(\alpha=-6\) gives \(5-12+\beta=0\), so \(\beta=7\). Hence \((\alpha,\beta)=(-6,7)\).

Explanation. For a line with direction \(\vec{b}=2\hat{i}+\hat{j}-2\hat{k}\) and a plane with normal \(\vec{n}=\hat{i}+\hat{j}\), the angle satisfies \(\sin\theta=\dfrac{|\vec{b}\cdot\vec{n}|}{|\vec{b}|\,|\vec{n}|}\). Here \(\vec{b}\cdot\vec{n}=2+1=3\), \(|\vec{b}|=3\) and \(|\vec{n}|=\sqrt2\), so \(\sin\theta=\dfrac{3}{3\sqrt2}=\dfrac{1}{\sqrt2}\), giving \(\theta=45^{\circ}\).

Explanation. A general point on the line is \((6-t,\,-1,\,-3+4t)\). Substituting into the plane \(x+y-z=3\): \((6-t)+(-1)-(-3+4t)=3\), which simplifies to \(8-5t=3\), so \(t=1\). The intersection point is \((5,-1,1)\).

Explanation. The distance from the origin to the plane \(ax+by+cz+d=0\) is \(\dfrac{|d|}{\sqrt{a^{2}+b^{2}+c^{2}}}\). Here it equals \(\dfrac{|7|}{\sqrt{9+36+4}}=\dfrac{7}{\sqrt{49}}=1\).

Explanation. Scale the first plane to share the second's normal: \(2x+4y+6z+14=0\). The two parallel planes \(2x+4y+6z+14=0\) and \(2x+4y+6z+7=0\) are separated by \(\dfrac{|14-7|}{\sqrt{4+16+36}}=\dfrac{7}{\sqrt{56}}=\dfrac{7}{2\sqrt{14}}\). Rationalising, \(\dfrac{7}{2\sqrt{14}}=\dfrac{\sqrt{14}}{4}=\dfrac{\sqrt{7}}{2\sqrt{2}}\).

Explanation. Direction cosines satisfy \(l^{2}+m^{2}+n^{2}=1\). Substituting \(\dfrac{1}{c^{2}}+\dfrac{1}{c^{2}}+\dfrac{1}{c^{2}}=1\) gives \(\dfrac{3}{c^{2}}=1\), so \(c^{2}=3\) and \(c=\pm\sqrt{3}\).

Explanation. Setting \(t=0\) gives the point \((1,-2,-1)\). Setting \(t=1\) gives \((1,\,-2+6,\,-1-1)=(1,4,-2)\). Both points have \(x=1\) because the direction vector has no \(\hat{i}\) component, so the line passes through \((1,-2,-1)\) and \((1,4,-2)\).

Explanation. The distance from \((1,1,1)\) to the origin is \(\sqrt3\). Its distance to the plane is \(\dfrac{|1+1+1+k|}{\sqrt3}=\dfrac{|3+k|}{\sqrt3}\). The condition \(\sqrt3=\tfrac12\cdot\dfrac{|3+k|}{\sqrt3}\) gives \(|3+k|=6\), so \(3+k=\pm6\), yielding \(k=3\) or \(k=-9\).

Explanation. Parallel planes have proportional normals: \(\dfrac{2}{4}=\dfrac{-\lambda}{1}=\dfrac{1}{-\mu}\). The common ratio is \(\tfrac12\). From \(-\lambda=\tfrac12\), \(\lambda=-\tfrac12\); from \(\dfrac{1}{-\mu}=\tfrac12\), \(-\mu=2\), so \(\mu=-2\). Hence \(\lambda=-\tfrac12,\ \mu=-2\).

Explanation. Write the plane as \(2x+3y+\lambda z-1=0\). The perpendicular distance from the origin is \(\dfrac{|-1|}{\sqrt{4+9+\lambda^{2}}}=\dfrac{1}{\sqrt{13+\lambda^{2}}}\). Setting this equal to \(\tfrac15\) gives \(\sqrt{13+\lambda^{2}}=5\), so \(13+\lambda^{2}=25\) and \(\lambda^{2}=12\). Since \(\lambda>0\), \(\lambda=2\sqrt{3}\).