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  • Need solution for RD sharma maths class 12 chapter 22 Algebra of Vector exercise 22.5 question 9

Need solution for RD sharma maths class 12 chapter 22 Algebra of Vector exercise 22.5 question 9

Answers (3)

the magnitude of AB and AC is equal

Hence the points 2\hat{i},-\hat{i}-4\hat{j} and -\hat{i}+4\hat{j}  form an isosceles triangle

Hint: the magnitude of \left |\vec{a} \right |=\sqrt{x^{2}+y^{2}}

Position vector \vec{AB} is given by the  \vec{AB} = position vector of B - position vector of A

Given:  the points A,B,C with position vectors \vec{a},\vec{b},\vec{c} respectively

Also,

\vec{a}=2\hat{i}\\ \vec{b}=-\hat{i}-4\hat{j}\\ \vec{c}=-\hat{i}+4\hat{j}

Solution:

Now we find the position vector \vec{AB}

Then,

\vec{AB}=\vec{b}-\vec{a}\\ =\left (-\hat{i}-4\hat{j} \right )-2\hat{i}\\ =-\hat{i}-4\hat{j}-2\hat{i}\\ \vec{AB}=-3\hat{3}-4\hat{j}

Now

\left |\vec{AB} \right |=\sqrt{x^{2}+y^{2}}

Here x=-3 and y= -4

=\sqrt{(-3)^{2}+(-4)^{2}}\\ =\sqrt{9+16}=\sqrt{25}\\ =5

\vec{BC}=Position vector of C – Position vector of B

=\left ( -\hat{i}+4\hat{j} \right )-\left ( -\hat{i}-4\hat{j} \right )\\ = -\hat{i}+4\hat{j}+\hat{i}+4\hat{j} \\ =0\hat{i}+8\hat{j}\\ =8\hat{j}

\vec{AC}=Position vector of C – position vector of A

=\left ( -\hat{i}+4\hat{j} \right )-\left ( 2\hat{i} \right )\\ = -\hat{i}+4\hat{j}-\hat{j} \\ =-3\hat{i}+4\hat{j}\\

Now

\left |\vec{AC} \right |=\sqrt{x^{2}+y^{2}}

Here x=-3 and y=4

=\sqrt{(-3)^{2}+(-4)^{2}}\\ =\sqrt{9+16}=\sqrt{25}\\ =5

Since, the magnitude of AB and AC are equal

Hence, the points 2\hat{i},-\hat{i}-4\hat{j} and -\hat{i}+4\hat{j} form an isosceles triangle

Posted by

Info Expert 29

View full answer

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the magnitude of AB and AC is equal

Hence the points 2\hat{i},-\hat{i}-4\hat{j} and -\hat{i}+4\hat{j}  form an isosceles triangle

Hint: the magnitude of \left |\vec{a} \right |=\sqrt{x^{2}+y^{2}}

Position vector \vec{AB} is given by the  \vec{AB} = position vector of B - position vector of A

Given:  the points A,B,C with position vectors \vec{a},\vec{b},\vec{c} respectively

Also,

\vec{a}=2\hat{i}\\ \vec{b}=-\hat{i}-4\hat{j}\\ \vec{c}=-\hat{i}+4\hat{j}

Solution:

Now we find position vector \vec{AB}

Then,

\vec{AB}=\vec{b}-\vec{a}\\ =\left (-\hat{i}-4\hat{j} \right )-2\hat{i}\\ =-\hat{i}-4\hat{j}-2\hat{i}\\ \vec{AB}=-3\hat{3}-4\hat{j}

Now

\left |\vec{AB} \right |=\sqrt{x^{2}+y^{2}}

Here x=-3 and y= -4

=\sqrt{(-3)^{2}+(-4)^{2}}\\ =\sqrt{9+16}=\sqrt{25}\\ =5

\vec{BC}=Position vector of C – Position vector of B

=\left ( -\hat{i}+4\hat{j} \right )-\left ( -\hat{i}-4\hat{j} \right )\\ = -\hat{i}+4\hat{j}+\hat{i}+4\hat{j} \\ =0\hat{i}+8\hat{j}\\ =8\hat{j}

\vec{AC}=Position vector of C – position vector of A

=\left ( -\hat{i}+4\hat{j} \right )-\left ( 2\hat{i} \right )\\ = -\hat{i}+4\hat{j}-\hat{j} \\ =-3\hat{i}+4\hat{j}\\

Now

\left |\vec{AC} \right |=\sqrt{x^{2}+y^{2}}

Here x=-3 and y=4

=\sqrt{(-3)^{2}+(-4)^{2}}\\ =\sqrt{9+16}=\sqrt{25}\\ =5

Since, the magnitude of AB and AC are equal

Hence, the points 2\hat{i},-\hat{i}-4\hat{j} and -\hat{i}+4\hat{j} form an isosceles triangle

Posted by

Info Expert 29

View full answer

the magnitude of AB and AC is equal

Hence the points 2\hat{i},-\hat{i}-4\hat{j} and -\hat{i}+4\hat{j}  form an isosceles triangle

Hint: the magnitude of \left |\vec{a} \right |=\sqrt{x^{2}+y^{2}}

Position vector \vec{AB} is given by the  \vec{AB} = position vector of B - position vector of A

Given:  the points A,B,C with position vectors \vec{a},\vec{b},\vec{c} respectively

Also,

\vec{a}=2\hat{i}\\ \vec{b}=-\hat{i}-4\hat{j}\\ \vec{c}=-\hat{i}+4\hat{j}

Solution:

Now we find position vector \vec{AB}

Then,

\vec{AB}=\vec{b}-\vec{a}\\ =\left (-\hat{i}-4\hat{j} \right )-2\hat{i}\\ =-\hat{i}-4\hat{j}-2\hat{i}\\ \vec{AB}=-3\hat{3}-4\hat{j}

Now

\left |\vec{AB} \right |=\sqrt{x^{2}+y^{2}}

Here x=-3 and y= -4

=\sqrt{(-3)^{2}+(-4)^{2}}\\ =\sqrt{9+16}=\sqrt{25}\\ =5

\vec{BC}=Position vector of C – Position vector of B

=\left ( -\hat{i}+4\hat{j} \right )-\left ( -\hat{i}-4\hat{j} \right )\\ = -\hat{i}+4\hat{j}+\hat{i}+4\hat{j} \\ =0\hat{i}+8\hat{j}\\ =8\hat{j}

\vec{AC}=Position vector of C – position vector of A

=\left ( -\hat{i}+4\hat{j} \right )-\left ( 2\hat{i} \right )\\ = -\hat{i}+4\hat{j}-\hat{j} \\ =-3\hat{i}+4\hat{j}\\

Now

\left |\vec{AC} \right |=\sqrt{x^{2}+y^{2}}

Here x=-3 and y=4

=\sqrt{(-3)^{2}+(-4)^{2}}\\ =\sqrt{9+16}=\sqrt{25}\\ =5

Since, the magnitude of AB and AC are equal

Hence, the points 2\hat{i},-\hat{i}-4\hat{j} and -\hat{i}+4\hat{j} form an isosceles triangle

Posted by

Info Expert 29

View full answer

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