NCERT Solutions · Class 12 Mathematics Vector Algebra

Let \(\displaystyle \vec r\) be equally inclined to OX, OY and OZ, making the same angle \(\displaystyle \alpha\) with each. By definition its direction cosines are \[l=\cos\alpha,\qquad m=\cos\alpha,\qquad n=\cos\alpha,\] so \(\displaystyle l=m=n\).The direction cosines of any vector satisfy \(\displaystyle l^{2}+m^{2}+n^{2}=1\). Hence \[3\cos^{2}\alpha=1\ \Longrightarrow\ \cos^{2}\alpha=\frac{1}{3}\ \Longrightarrow\ \cos\alpha=\pm\frac{1}{\sqrt{3}}.\]Both signs must be kept: taking the square root of \(\displaystyle \cos^{2}\alpha\) admits an acute angle \(\displaystyle \alpha\) (for a vector in the direction of \(\displaystyle \hat i+\hat j+\hat k\)) and an obtuse one (for the oppositely directed vector \(\displaystyle -\left(\hat i+\hat j+\hat k\right)\)); in each case the three cosines are equal to one another.Therefore the direction cosines are \[(l,m,n)=\pm\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right).\]

A vector perpendicular to both \(\displaystyle \vec a\) and \(\displaystyle \vec b\) must be parallel to \(\displaystyle \vec a\times\vec b\), so write \(\displaystyle \vec d=\lambda\left(\vec a\times\vec b\right)\) and fix \(\displaystyle \lambda\) from the second condition.\[\vec a\times\vec b=\left|\begin{array}{rrr}\hat i & \hat j & \hat k\\ 1 & 4 & 2\\ 3 & -2 & 7\end{array}\right|=\hat i\left(28+4\right)-\hat j\left(7-6\right)+\hat k\left(-2-12\right)=32\hat i-\hat j-14\hat k.\]So \(\displaystyle \vec d=\lambda\left(32\hat i-\hat j-14\hat k\right)\). Imposing \(\displaystyle \vec c\cdot\vec d=15\) with \(\displaystyle \vec c=2\hat i-\hat j+4\hat k\): \[\vec c\cdot\vec d=\lambda\left(2\cdot 32+(-1)(-1)+4(-14)\right)=\lambda\left(64+1-56\right)=9\lambda.\] \[9\lambda=15\ \Longrightarrow\ \lambda=\frac{5}{3}.\]Hence \[\vec d=\frac{5}{3}\left(32\hat i-\hat j-14\hat k\right)=\frac{1}{3}\left(160\hat i-5\hat j-70\hat k\right).\]

Sum of the two vectors: \[\vec s=\left(2\hat i+4\hat j-5\hat k\right)+\left(\lambda\hat i+2\hat j+3\hat k\right)=(2+\lambda)\hat i+6\hat j-2\hat k.\] \[|\vec s|=\sqrt{(2+\lambda)^{2}+36+4}=\sqrt{(\lambda+2)^{2}+40}.\]Unit vector along \(\displaystyle \vec s\) is \(\displaystyle \hat s=\dfrac{\vec s}{|\vec s|}\), and the given condition is \(\displaystyle \left(\hat i+\hat j+\hat k\right)\cdot\hat s=1\): \[\frac{(2+\lambda)+6-2}{\sqrt{(\lambda+2)^{2}+40}}=1\ \Longrightarrow\ \lambda+6=\sqrt{(\lambda+2)^{2}+40}.\]Squaring (legitimate only if \(\displaystyle \lambda+6\ge 0\), which must be checked at the end): \[\lambda^{2}+12\lambda+36=\lambda^{2}+4\lambda+4+40\ \Longrightarrow\ 8\lambda=8\ \Longrightarrow\ \lambda=1.\]Check: with \(\displaystyle \lambda=1\), \(\displaystyle \lambda+6=7>0\) and \(\displaystyle \sqrt{3^{2}+40}=\sqrt{49}=7\), so the equation is satisfied and no root has been introduced by squaring.\(\displaystyle \lambda=1\).

By definition of the scalar product, \(\displaystyle \vec a\cdot\vec b=|\vec a||\vec b|\cos\theta\). Since \(\displaystyle |\vec a|\ge 0\) and \(\displaystyle |\vec b|\ge 0\), \[\vec a\cdot\vec b\ge 0\iff\cos\theta\ge 0.\]For the angle between two vectors, \(\displaystyle 0\le\theta\le\pi\), and on this range \(\displaystyle \cos\theta\ge 0\) exactly for \(\displaystyle 0\le\theta\le\frac{\pi}{2}\) — the endpoints must be included, since \(\displaystyle \cos 0=1>0\) and \(\displaystyle \cos\frac{\pi}{2}=0\), and \(\displaystyle \ge\) admits the value 0. That is why (A), with its strict inequalities, is not the answer.Answer: (B) \(\displaystyle 0\le\theta\le\dfrac{\pi}{2}\).

Since \(\displaystyle \vec a\) and \(\displaystyle \vec b\) are unit vectors, \(\displaystyle |\vec a|=|\vec b|=1\) and \(\displaystyle \vec a\cdot\vec b=\cos\theta\). Then \[|\vec a+\vec b|^{2}=(\vec a+\vec b)\cdot(\vec a+\vec b)=|\vec a|^{2}+|\vec b|^{2}+2\,\vec a\cdot\vec b=1+1+2\cos\theta=2+2\cos\theta.\]For \(\displaystyle \vec a+\vec b\) to be a unit vector, \(\displaystyle |\vec a+\vec b|^{2}=1\): \[2+2\cos\theta=1\ \Longrightarrow\ \cos\theta=-\frac{1}{2}\ \Longrightarrow\ \theta=\frac{2\pi}{3}\quad(0\le\theta\le\pi).\]Answer: (D) \(\displaystyle \theta=\dfrac{2\pi}{3}\).

Use \(\displaystyle \hat i\times\hat j=\hat k,\ \hat j\times\hat k=\hat i,\ \hat k\times\hat i=\hat j\), and note that reversing the order reverses the sign, so \(\displaystyle \hat i\times\hat k=-\hat j\). Also \(\displaystyle \hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1\).\[\hat i\cdot(\hat j\times\hat k)=\hat i\cdot\hat i=1,\] \[\hat j\cdot(\hat i\times\hat k)=\hat j\cdot(-\hat j)=-1,\] \[\hat k\cdot(\hat i\times\hat j)=\hat k\cdot\hat k=1.\]The middle term is the one to watch: \(\displaystyle \hat i\times\hat k\) is taken in the reverse of the cyclic order \(\displaystyle \hat k\times\hat i=\hat j\), so it equals \(\displaystyle -\hat j\).Sum \(\displaystyle =1+(-1)+1=1\).Answer: (C) 1.

For the angle \(\displaystyle \theta\) between two vectors, \(\displaystyle 0\le\theta\le\pi\), and \[|\vec a\cdot\vec b|=|\vec a||\vec b||\cos\theta|,\qquad |\vec a\times\vec b|=|\vec a||\vec b|\sin\theta,\] where \(\displaystyle \sin\theta\ge 0\) on this range, which is why no modulus is needed on the sine.Assuming \(\displaystyle \vec a,\vec b\) are non-zero, equating the two and cancelling \(\displaystyle |\vec a||\vec b|\neq 0\): \[|\cos\theta|=\sin\theta.\] If \(\displaystyle \cos\theta=0\) this forces \(\displaystyle \sin\theta=0\) too, which is impossible; so \(\displaystyle \cos\theta\neq 0\) and we may divide: \[|\tan\theta|=1\ \Longrightarrow\ \theta=\frac{\pi}{4}\ \text{or}\ \theta=\frac{3\pi}{4}.\]Of these only \(\displaystyle \frac{\pi}{4}\) appears among the options.Answer: (B) \(\displaystyle \dfrac{\pi}{4}\).