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# A image of the point P with position vector $7i - j + 2k$ in the line whose vector equation is $\overline r - 9i + 5j + 5k + 4(i - 3j + 5k)$has the position vector.

Last updated date: 14th Jun 2024
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Hint: Vectors refers to lines that represent both direction and magnitude (size) consequently if an object move in more than one direction or more than one force acts upon an object equivalently than we can add the vectors to find a resultant displacement or resultant force that acts on it.

Position vector: - The position vector is given with respect to the origin O$\left( {o,o,o} \right)$where the coordinate of point 9sayp) is $p(x,y,z)$ is given as:-
$\overline {OP} = \sqrt {{x^2} + {y^2} + {z^2}}$
Components of the vector
Let us assume an xyz coordinate plane and unit vector i, j and k are defined across $x,y,z$respectively. Then we can represent any vector in the components from like
$r = xi + yi + zk$
Length of vector or magnitude of the vector is defined as $\sqrt {x{}^2 + {y^2} + {z^2}}$
$x,y$ and $z$ are scalar components of vector r
$xi,yi,zk$ are called the vector components
$x,y,z$ are termed as rectangular components.
Therefore

Let $(\alpha ,\beta ,\gamma )$be the position vectors
$\therefore \dfrac{{\alpha + \gamma }}{2}\mathop j\limits^{} + \dfrac{{B - 1}}{2}\mathop j\limits^{} + \dfrac{{y + 2}}{2}\mathop k\limits^{}$
$= (9\mathop j\limits^{} + 5\mathop j\limits^{} + 5\mathop k\limits^{} ) + \alpha (j + 3\mathop j\limits^{} + 5\mathop k\limits^{} )$
Comparing
We get
$(\dfrac{{\alpha + 7}}{2}) = 9 + \alpha$
$\alpha - 7 = 18 + 2\alpha$
$\alpha = 18 - 7 + 2\alpha$
$\alpha = 11 + 2\alpha ...........(1)$
$(\dfrac{{\beta - 1}}{2}) = 5 + 3\alpha$
$\beta - 1 = 10 + 6\alpha$
$\beta = 11 + 6\alpha ..........(2)$
$(\dfrac{{\lambda + 2}}{2}) = 5 + 5\alpha$
$\lambda + 2 = 10 + 10\alpha$
$\lambda = 8 + 10\alpha ..........(3)$
$\dfrac{{x - 9}}{1} = \dfrac{{y - 5}}{3} = \dfrac{{z - 5}}{5}$
Given point
$p(7, - 1,2)\,\,\,\,Q(a,b,c)$
Direction of PQ$= (a - 7,b - 1,c - 2)$
$(\dfrac{{\beta - 1}}{2}) = 5 + 3\alpha$
$\beta - 1 = 10 + 6\alpha$
$\beta = 11 + 6\alpha ..........(2)$
$(\dfrac{{\lambda + 2}}{2}) = 5 + 5\alpha$
$\lambda + 2 = 10 + 10\alpha$
$\lambda = 8 + 10\alpha ..........(3)$
PQ and given line are 1 (perpendicular) so
${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0$
$\Rightarrow (\alpha - 7) + 3(\beta + 1) + 5(\alpha - 2) = 0$
$\alpha + 3\beta + 5\lambda = 14$
$\therefore 11 + 2\alpha + 3(11 + 6\alpha ) + 5(8 + 10\alpha ) = 14$
$\therefore 70 + 70\alpha = 0$
$\lambda = - 1$
Put this value in $e{q^n}$ 91) (2) & (3)
We get
$\therefore (\alpha ,\beta ,\lambda ) = (9,5, - 2)$
Hence the image of the point P are$(9,5, - 2)$

Note:
Commutative law $= A + B = B + A$
Associative law $= A(B + C) = (A + B) + C$
Dot product $\to P.Q = \left| P \right|\left| Q \right|\operatorname{Cos} \theta$
Cross product $\to P \times Q = \left| P \right|\left| Q \right|\operatorname{Sin} \theta$
$li + nj + nk = (\operatorname{Cos} \theta x)i + (\operatorname{Cos} \theta y)j + (\operatorname{Cos} \theta z)k$
Is the unit vector in the direction of the vector and $\theta x,\theta y,\theta z$ are the angles which the vector makes with $x,y$ and z axis.