Introduction to vector algebra - class-XI

Using the identity |a x b|^2 + (a . b)^2 = |a|^2 * |b|^2, we have 144 = 4^2 * |b|^2. Thus, 144 = 16 * |b|^2, which means |b|^2 = 9, so |b| = 3.

Displacement,Centrifugal force and Gravitation are vectors.
Work $W=\overrightarrow { F } \cdot \overrightarrow { S } $ a scalar.
Hence, option B.

Temperature,Density,Mass are clearly scalars.
Whereas in case of weight, it acts always downwards beacuse of
accelration due to gravity.
Hence, option D.

For a given unit vector , Magnitude and direction are constant (fixed).
since, a vector cannot have two directions.
Hence, option C.

 A quantity that has magnitude as well as direction is called a vector.  

We know that the symbolically is  $\leftrightarrow$

The vector is visual representation of a physical quantity that which has the both direction and the magnitude.

The vector is visual representation of a physical quantity that which has the both direction and the magnitude.

$\vec u + \vec v$ represents vector.

The scalar triple product [a b c] is a scalar value. The expression [a x b x c] = 2[a b c]^2 is not a standard vector identity and is dimensionally inconsistent.

$\\Let\>A\>(2\hat{i}-\hat{j}+\hat{k}),\>B(\hat{i}-3\hat{j}-5\hat{k})\>and\\C(3\hat{i}-4\hat{j}-4\hat{k})\\then\\\overrightarrow{AB}=-\hat{i}-2\hat{j}-6\hat{k}\\\therefore\>|\overrightarrow{AB}|=\sqrt{1+4+36}=\sqrt{41}\\\overrightarrow{BC}=2\hat{i}-\hat{j}+\hat{k}\\\therefore\>|\overrightarrow{BC}|=\sqrt{4+1+1}=\sqrt{6}\\\overrightarrow{CA}=-\hat{i}+3\hat{j}+5\hat{k}\\\therefore\>|\overrightarrow{CA}|=\sqrt{1+9+25}=\sqrt{35}\\clearly\>\>\>|\overrightarrow{BC}|^2+|\overrightarrow{CA}|^2=|\overrightarrow{AB}|^2\\\therefore\>Triangle\>is\>a\>right\>angled\>triangle$

$\begin{array}{l} \left| { \begin{array} { *{ 20 }{ c } }{ { { \sec   }^{ 2 } }A } & 1 & 1 \ 1 & { { { \sec   }^{ 2 } }B } & 1 \ 1 & 1 & { { { \sec   }^{ 2 } }C } \end{array} } \right| =0 \\ { C _{ 1 } }\to { C _{ 1 } }-{ C _{ 2 } }, , , , , , , , , { C _{ 2 } }\to { C _{ 2 } }-{ C _{ 3 } } \\ \left| { \begin{array} { *{ 20 }{ c } }{ { { \tan   }^{ 2 } }A } & 0 & 1 \ { -{ { \tan   }^{ 2 } }B } & { { { \tan   }^{ 2 } }B } & 1 \ 0 & { -{ { \tan   }^{ 2 } }C } & { { { \sec   }^{ 2 } }C } \end{array} } \right| =0 \\ { \tan ^{ 2 }  }A\left[ { { { \tan   }^{ 2 } }B{ { \sec   }^{ 2 } }C+{ { \tan   }^{ 2 } }C } \right] +{ \tan ^{ 2 }  }B{ \tan ^{ 2 }  }C=0 \ \\dfrac { { { { \sec   }^{ 2 } }C } }{ { { { \tan   }^{ 2 } }C } } +{ \cot ^{ 2 }  }B+{ \cot ^{ 2 }  }A=0 \ \\cos  e{ c^{ 2 } }C+{ \cot ^{ 2 }  }B+{ \cot ^{ 2 }  }A=0 \\ { \cot ^{ 2 }  }A+{ \cot ^{ 2 }  }B+{ \cot ^{ 2 }  }C=-1 \\ Hence,, the, option, D, is, , the, correct, answer. \end{array}$

Given the conditions that (a+b) is perpendicular to c and the projection of b on a is a, one can solve for the components of b. The calculation leads to |b| = sqrt(22).

Let$b=xi+yj+zk$

If a vector of magnitude 6 is inclined at equal angles to the axes, its components are equal (x=y=z). Thus, V = k(i + j + k). Since |V| = 6, k * sqrt(3) = 6, so k = 6/sqrt(3) = 2 * sqrt(3).

Given sum(ai) = 0, we have |sum(ai)|^2 = 0. Expanding this, sum(|ai|^2) + 2 * sum(ai . aj) = 0. Since |ai| = 1, sum(1) + 2 * sum(ai . aj) = 0, so n + 2 * sum(ai . aj) = 0, which gives sum(ai . aj) = -n/2.

Given $ |\vec{a}+\vec{b}| = \sqrt{3}, $

As, zero vector represents a point.
Direction is indeterminate.
Hence, option C.

length of vector $a\hat { i } +b\hat { j } +c\hat { k } $ from origin is $\sqrt { a^2+b^2+c^2 } $ 

$|\vec{z} + (\vec{z} \times \vec{x}) | = | \vec{y}|^2 ,,,,,\Rightarrow$
$|\vec{z}|^2+|\vec{z}|^2 |\vec{x}|^2 ,sin^2,\theta =1$
$\displaystyle \Rightarrow ,|z| = \frac{1}{\sqrt {1 + sin^2,\theta}} = \frac{2}{\sqrt 7} \Rightarrow sin,\theta = \frac{\sqrt 3}{2}$