Solution $$ \begin{array}{l} \vec{a}=\hat{\imath}+2 \hat{\jmath}+3 k \\ \mathrm{I} \overrightarrow{\mathrm{I}}=\sqrt{1} 4 \end{array} $$ Unit vector along $\overrightarrow{a_{1}...
![\[\text { de direction cosines of the vector } \overrightarrow{\mathrm{a}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8244f35a251c243e63649ced6e4116bb_l3.png "Rendered by QuickLaTeX.com")
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Rs aggarwal Class 12 Chapter Product of three vectors Exercise 25 B
Solution $$ \begin{array}{l} \vec{a}=\hat{\imath}+2 \hat{\jmath}+3 k \\ \mathrm{I} \overrightarrow{\mathrm{I}}=\sqrt{1} 4 \end{array} $$ Unit vector along $\overrightarrow{a_{1}...
![\[\begin{array}{l} \text { (d) } \vec{P}=\hat{\imath}+5 \hat{\jmath}+4 \hat{k}\\ \widehat{D} \vec{Q}=4 \hat{\imath}+\hat{\jmath}-2 \hat{k}\\ \text { BH) } \mid \hat{0}=3 \hat{\imath}-4 \hat{\jmath}-6 \hat{k}\\ \text { Direction ratios of } \forall H) \overrightarrow{\mathbb{Q}} \text { are }(3,-4,-6) \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7bc40016d17abb9221384af22f463652_l3.png "Rendered by QuickLaTeX.com")
Solution If $P(1,5,4)$ and $Q(4,1,-2)$ be the position vectors of two points $P$ and $Q$, find the direction ratios of $\overrightarrow{\mathrm{PO}}$
Find the direction cosines of a vector which is equally inclined to the 
and 
– axis.
Solution Let $\alpha, \beta, \gamma$ be the angles made by the vector with $\mathrm{x}$ - axis, $\mathrm{y}$ axis and z- axis respectively Given: $\alpha=\beta=\gamma$ Direction cosines of vector...
![\[\text { When does }|\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}|_{\text {hold? }}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ee1616736136c9e520b861aff6f9dd7f_l3.png "Rendered by QuickLaTeX.com")
Solution Given: $\mathrm{I} \vec{a}+\vec{b} \mathrm{I}=\mathrm{I} \vec{a} \mathrm{I}+\mathrm{I} \vec{b} \mathrm{I}$ $$ \begin{array}{l} (\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=\mathrm{I}...
![\[\text { If } \theta \text { is the angle between } \overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{b}}, \text { and }|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|_{\text {then what is the value of } \theta \text { ? }}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f9c0d3ee62bcefae3697d380147d92c4_l3.png "Rendered by QuickLaTeX.com")
Solution $\theta$ is angle between $\vec{a}$ and $\vec{b}$ Given: $\mathrm{I} \vec{a} \cdot \vec{b} \mathrm{I}=\mathrm{I} \vec{a} \times \vec{b} \mathrm{I}$ $\mathrm{I} \vec{a} \mathrm{I} \mathrm{I}...
![\[\text { If } \overrightarrow{\mathrm{a}}=(2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+27 \hat{\mathrm{k}})_{\text {and }} \overrightarrow{\mathrm{b}}=(\hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}+\mu \hat{\mathrm{k}})_{\text {are such that }} \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{0} \text { then find the values }\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-57e65bede8831b40c17461115dddd0ad_l3.png "Rendered by QuickLaTeX.com")
of 
and 
.
Solution $\vec{a}=2 \hat{\imath}+6 \hat{\jmath}+27 k$ $\vec{b}=\hat{\imath}+\lambda \hat{\jmath}+\mu k$ $\vec{a} \times \vec{b}=() \hat{\imath}-() \hat{\jmath}+() k$ $\vec{a} \times \vec{b}=0$ $2...
Find the volume of the parallelepiped whose edges are represented by the vectors 
Solution $$ \begin{array}{l} \vec{a}=2 \hat{\imath}-3 \hat{\jmath}+4 k \\ \vec{b}=\hat{\imath}+2 \hat{\jmath}-k \\ \vec{c}=3 \hat{\imath}-2 \hat{\jmath}+2 k \end{array} $$ To find the volume of...
Write the value of ![\[\hat{\mathrm{i}} \cdot(\hat{\mathrm{j}} \times \hat{\mathrm{k}})+\hat{\mathrm{j}} \cdot(\hat{\mathrm{i}} \times \hat{\mathrm{k}})+\hat{\mathrm{k}} \cdot(\hat{\mathrm{i}} \times \hat{\mathrm{j}})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1ba1063a7661057d24a42c3bf4d427ba_l3.png "Rendered by QuickLaTeX.com")
Solution $\hat{\imath} \cdot(\hat{\jmath} \times k)+\hat{\jmath} \cdot(\hat{\imath} \times k)+k \cdot(\hat{\imath} \times \hat{\jmath})$ We know that $\left[\begin{array}{lll}\hat{\imath} &...
![\[\text { en the vectors } \overrightarrow{\mathrm{a}}=(\hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}+3 \hat{\mathrm{k}})_{\text {and }} \overrightarrow{\mathrm{b}}=(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+9 \hat{\mathrm{k}})_{\text {are parallel. }}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6920759d4fa589725b60c82132455ae8_l3.png "Rendered by QuickLaTeX.com")
$$ \text { en the vectors } \overrightarrow{\mathrm{a}}=(\hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}+3 \hat{\mathrm{k}})_{\text {and }} \overrightarrow{\mathrm{b}}=(3 \hat{\mathrm{i}}+2...
What conclusion can you draw about vectors 
and 
when 
and 
Answer: Given: $\vec{a} \times \vec{b}=0$ and $\vec{a} \cdot \vec{b}=0$ If both cross and dot product of two vectors is zero, then it means that products are independent of angle between them. Then...
Find the angle between two vectors and with magnitudes 1 and 2 respectively, when 
Solution $\mathrm{I} \overrightarrow{\mathrm{I}}=1$ $\mathrm{I} \overrightarrow{\mathrm{b}} \mathrm{I}=2$ $\mathrm{I} \vec{a} \times \vec{b} \mathrm{I}=\sqrt{3}$ $\mathrm{I} \vec{a} \mathrm{I}...
![\[\text { If } \vec{a}=(\hat{i}-7 \hat{j}+7 \hat{k})_{\text {and }} \vec{b}=(3 \hat{i}-2 \hat{j}+2 \hat{k})_{\text {then find }}|\vec{a} \times \vec{b}|\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1de7056c74786d7709a78838d96667b1_l3.png "Rendered by QuickLaTeX.com")
Solution $\vec{a}=\hat{\imath}-7 \hat{\jmath}+7 \hat{k}$ $\vec{b}=3 \hat{\imath}-2 \hat{\jmath}+2 k$ $\vec{a} \times \vec{b}=(-14+14) \hat{\imath}-(2-21) \hat{\jmath}+(-2+21) k$ $\vec{a} \times...
Write the angle between two vectors and with magnitudes 
and 2 respectively having 
Solution $\mathrm{I} \overrightarrow{\mathrm{I}} \mathrm{I}=\sqrt{3}$ $\mathrm{I} \overrightarrow{\mathrm{b}} \mathrm{I}=2$ $\vec{a} \cdot \vec{b}=\sqrt{6}$ $\vec{a} \cdot \vec{b}=\mathrm{I} \vec{a}...
![\[\text { Write the projection of the vector }(\hat{i}-\hat{j}) \text { on the vector }(\hat{i}+\hat{j}) .\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a23c5a53b84edf8638e83f8acce20522_l3.png "Rendered by QuickLaTeX.com")
Solution $$ \begin{array}{l} \vec{a}=\hat{\imath}-\hat{\jmath} \\ \vec{b}=\hat{\imath}+\hat{\jmath} \\ \vec{a} \cdot \vec{b}=1-1=0 \\ \mathrm{I} \vec{a} \mathrm{I}=\sqrt{2} \\ \mathrm{I} \vec{b}...
![\[\text { If } \vec{a}=(\hat{i}+\hat{j}+\hat{k}), \vec{b}=(4 \hat{i}-2 \hat{j}+3 \hat{k})_{\text {and }} \vec{c}=(\hat{i}-2 \hat{j}+\hat{k}), \text { find a vector of magnitude } 6\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7b36e973b1ab4f6dc2f92337385ffb23_l3.png "Rendered by QuickLaTeX.com")
units which is parallel to the vector 
.
Solution $\vec{a}=\hat{\imath}+\hat{\jmath}+k$ $\vec{b}=4 \hat{\imath}-2 \hat{\jmath}+3 k$ $\vec{c}=\hat{\imath}-2 \hat{\jmath}+k$ $2 \vec{a}-\vec{b}+3 \vec{c}=2(\hat{\imath}+\hat{\jmath}+k)-(4...
![\[\text { e the vector of magnitude } 15 \text { units in the direction of vector }(\hat{i}-2 \hat{j}+2 \hat{k})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-84b18540a8494956188eb8b12e53934a_l3.png "Rendered by QuickLaTeX.com")
Solution $$ \begin{array}{l} \vec{a}=\hat{\imath}-2 \hat{\jmath}+2 k \\ \mathrm{I} \vec{a} \mathrm{I}=3 \end{array} $$ Unit vector in direction of required vector is $\left(\mathbf{I} \hat{\imath}+{...
![\[\text { If } \overrightarrow{\mathrm{a}}=(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}), \overrightarrow{\mathrm{b}}=(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})_{\text {a } \mathrm{d}} \overrightarrow{\mathrm{c}}=(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})_{\text {are such that }}(\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}})_{\text {is }}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e8494195f37058df991aad951c942848_l3.png "Rendered by QuickLaTeX.com")
perpendicular to 
then find the value of 
.
Solution $\vec{a}=2 \hat{\imath}+2 \hat{\jmath}+3 k$ $\vec{b}=-\hat{\imath}+2 \hat{\jmath}+k$ $$ \begin{array}{l} \vec{c}=3 \hat{\imath}+\hat{\jmath} \\ \vec{a}+\lambda \vec{b}=(\hat{\imath}+2...
![\[\text { Find } \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})_{\text {when }} \overrightarrow{\mathrm{a}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}), \overrightarrow{\mathrm{b}}=(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})_{\text {and }} \overrightarrow{\mathrm{c}}=(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7406c2978bc2544d97fa573cbd077db9_l3.png "Rendered by QuickLaTeX.com")
Solutioin $\vec{a}=2 \hat{\imath}+\hat{\jmath}+3 k$ $\vec{b}=-\hat{\imath}+2 \hat{\jmath}+k$ $\vec{c}=3 \hat{\imath}+\hat{\jmath}+2 k$ $\vec{a} \cdot \vec{b} \times...
![\[\text { te the projection of the vector } (7 \hat{i}+\hat{j}-4 \hat{k}) \text { on the vector }(2 \hat{i}+6 \hat{j}+3 \hat{k})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-89aa0cc829d2c9efecc4ea9942a150dd_l3.png "Rendered by QuickLaTeX.com")
Solution $$ \begin{array}{l} \vec{a}=7 \hat{\imath}+\hat{\jmath}-4 k \\ \vec{b}=2 \hat{\imath}+6 \hat{\jmath}+3 k \\ \vec{a} \cdot \vec{b}=8 \\ \mathrm{I} \vec{a} \mathrm{I}=\sqrt{66} \\ \mathrm{I}...
![\[\text { Write the projection of the vector }(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \text { along the vector } \hat{\mathrm{j}} \text {. }\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a278fc4cab1975e27ab7b277eb2349fb_l3.png "Rendered by QuickLaTeX.com")
Solution $$ \begin{array}{l} \vec{a}=\hat{\imath}+\hat{\jmath}+\not k \\ \vec{b}=\hat{\jmath} \end{array} $$ Projection of $\vec{a}$ on $\vec{b}$ is nothing but $\mathrm{I} \vec{a} \mathrm{I} \cos...
Find the sum of the vectors 
and 
Solution $\vec{a}=\hat{\imath}-2 \hat{\jmath}$ $\vec{b}=2 \hat{\imath}-3 \hat{\jmath}$ $\vec{c}=2 \hat{\imath}+3 \hat{k}$
![\[\text { Find the sum of the vectors } \overrightarrow{\mathrm{a}}=(\hat{\mathrm{i}}-3 \hat{\mathrm{k}}), \overrightarrow{\mathrm{b}}=(2 \hat{\mathrm{j}}-\hat{\mathrm{k}})_{\text {and }} \overrightarrow{\mathrm{c}}=(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aff5c3fe0957a4a555cbece3685a7597_l3.png "Rendered by QuickLaTeX.com")
Solution $\vec{a}=\hat{\imath}-3 k$ $\vec{b}=2 \hat{\jmath}-k$ $\vec{c}=2 \hat{\imath}-3 \hat{\jmath}+2 k$ $\vec{a}+\vec{b}+\vec{c}=3 \hat{\imath}-\hat{\jmath}-2 k$
![\[\text { If } \vec{a} \text { is a unit vector such that }(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=15, \text { find }|\vec{x}| \text {. }\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c11d4ceb6c44f39aa3e9e3aec1b7ef31_l3.png "Rendered by QuickLaTeX.com")
Solution $\mathrm{I} \vec{a} \mathrm{I}=1$ Given: $(\vec{x}-\vec{a})(\vec{x}+\vec{a})=15$ $\mathrm{I} \vec{x} \mathrm{I}^{2}-\mathrm{I} \vec{a} \mathrm{I}^{2}=15$ $\mathrm{I} \vec{x}...
If and are perpendicular vectors such that 
and 
, find the value of 
If $\vec{a}$ and $\vec{b}$ are perpendicular vectors such that $|\vec{a}+\vec{b}|=13$ and $|\vec{a}|=5$, find the value of $|\overrightarrow{\mathrm{b}}|$ Solution $$ \begin{array}{l} \mathrm{I}...
![\[\vec{a}=(\lambda \hat{i}+\hat{j}+4 \hat{k})_{\text {on }} \vec{b}=(2 \hat{i}+6 \hat{j}+3 \hat{k})_{\text {is } 4}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0dab27ad3c7020103c23d3e44f3c1eb0_l3.png "Rendered by QuickLaTeX.com")
units.
Solution $\vec{a}=\lambda \hat{\imath}+\hat{\jmath}+4 \hat{k}$ $$ \vec{b}=2 \hat{\imath}+6 \hat{\jmath}+3 \hat{k} $$ Since the projection of $\vec{a}$ on $\vec{b}$ is 4 units $\mathrm{I} \vec{a}...
![\[\overrightarrow{\mathrm{a}}=(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+9 \hat{\mathrm{k}})_{\text {and }} \overrightarrow{\mathrm{b}}=(\hat{\mathrm{i}}-2 \mathrm{p} \hat{\mathrm{j}}+3 \hat{\mathrm{k}})_{\text {are }}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-33af82b5565e885a5bbe135431306cc0_l3.png "Rendered by QuickLaTeX.com")
parallel.
Solution $$ \begin{array}{l} \vec{a}=3 \hat{\imath}+2 \hat{\jmath}+9 k \\ \vec{b}=\hat{\imath}-2 \mathrm{p} \hat{\jmath}+3 k \end{array} $$ Since $\vec{a}$ and $\vec{b}$ are parallel $$ \mathrm{I}...
Write the value of so that the vectors 
and 
are perpendicular to each other.
Solution $\vec{a}=2 \hat{\imath}+\lambda \hat{\jmath}+k$ $\vec{b}=\hat{\imath}-2 \hat{\jmath}+3 k$ If $\vec{a}$ is perpendicular to $\vec{b}$ Then $\vec{a} \cdot \vec{b}=0$ $$ \begin{array}{l} 2-2...
![\[\text { If } \vec{a}=x \hat{i}+2 \hat{j}-z \hat{k} \text { and } \vec{b}=3 \hat{i}-y \hat{j}+\hat{k} \text { are two equal vectors the } x+y+z=\text { ? }\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e9a0cd1b3f50e36cf71bda1d6f75a6c8_l3.png "Rendered by QuickLaTeX.com")
Solution $$ \begin{array}{l} \vec{a}=x \hat{\imath}+2 \hat{\jmath}-z k \\ \vec{b}=3 \hat{\imath}-y \hat{\jmath}+k \end{array} $$ Given $\vec{a}=\vec{b}$ $$ \mathrm{X}=3, \mathrm{y}=-2, \mathrm{z}=-1...